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Transcript of Stoichiometry
![Page 1: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/1.jpg)
StoichiometryPart 3: mass to moles
![Page 2: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/2.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
![Page 3: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/3.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g)
![Page 4: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/4.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)
![Page 5: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/5.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
![Page 6: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/6.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?
![Page 7: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/7.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
![Page 8: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/8.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
NH3(g) + O2(g) →NO(g) + H2O(g)
![Page 9: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/9.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
NH3(g) + O2(g) →NO(g) + H2O(g)
First, we balance the reaction.
![Page 10: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/10.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
NH3(g) + O2(g) →NO(g) + 3 H2O(g)
First, we balance the reaction.
![Page 11: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/11.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
2 NH3(g) + O2(g) →NO(g) + 3 H2O(g)
First, we balance the reaction.
![Page 12: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/12.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
2 NH3(g) + O2(g) →2 NO(g) + 3 H2O(g)
First, we balance the reaction.
![Page 13: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/13.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
2 NH3(g) + 2½ O2(g) →2 NO(g) + 3 H2O(g)
First, we balance the reaction.
![Page 14: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/14.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
2( 2 NH3(g) + 2½ O2(g) →2 NO(g) + 3 H2O(g) )
First, we balance the reaction.
![Page 15: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/15.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
First, we balance the reaction.
![Page 16: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/16.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.
![Page 17: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/17.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol)
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.
![Page 18: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/18.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.
![Page 19: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/19.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.
![Page 20: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/20.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.
![Page 21: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/21.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.
![Page 22: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/22.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.
![Page 23: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/23.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g)First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.
![Page 24: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/24.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.
![Page 25: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/25.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.Fourth, we write down the unknown values.
![Page 26: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/26.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.Fourth, we write down the unknown values.
![Page 27: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/27.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)?
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.Fourth, we write down the unknown values.
![Page 28: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/28.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.Fourth, we write down the unknown values.
![Page 29: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/29.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.Fourth, we write down the unknown values.Fifth, we plan our solution strategy.
![Page 30: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/30.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.Fourth, we write down the unknown values.Fifth, we plan our solution strategy.
1. Calculate the number of mols of NH3.
![Page 31: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/31.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.Fourth, we write down the unknown values.Fifth, we plan our solution strategy.
1. Calculate the number of mols of NH3.
⓵
![Page 32: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/32.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.Fourth, we write down the unknown values.Fifth, we plan our solution strategy.
1. Calculate the number of mols of NH3.2. Calculate the number of mols of NO.
⓵
![Page 33: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/33.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.Fourth, we write down the unknown values.Fifth, we plan our solution strategy.
1. Calculate the number of mols of NH3.2. Calculate the number of mols of NO.
⓵ ⓶
![Page 34: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/34.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.Fourth, we write down the unknown values.Fifth, we plan our solution strategy.
1. Calculate the number of mols of NH3.2. Calculate the number of mols of NO.3. Calculate the number of mols of H2O.
⓵ ⓶
![Page 35: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/35.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
First, we balance the reaction.Second, we write down the molar masses, Ṃ, in g/mol.Third, we write the known value.Fourth, we write down the unknown values.Fifth, we plan our solution strategy.
1. Calculate the number of mols of NH3.2. Calculate the number of mols of NO.3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 36: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/36.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
1. Calculate the number of mols of NH3.2. Calculate the number of mols of NO.3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 37: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/37.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
1. Calculate the number of mols of NH3.
2. Calculate the number of mols of NO.
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 38: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/38.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
1. Calculate the number of mols of NH3.
2. Calculate the number of mols of NO.
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 39: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/39.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia
2. Calculate the number of mols of NO.
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 40: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/40.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol)
2. Calculate the number of mols of NO.
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 41: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/41.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol)? ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 42: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/42.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.4? ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 43: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/43.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.4? ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 44: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/44.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.4? ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 45: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/45.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.4? ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 46: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/46.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.4? ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 47: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/47.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.4? ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 48: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/48.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.4? ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 49: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/49.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.448.4 ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 50: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/50.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.448.4 ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 51: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/51.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.448.4 ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.
⓵ ⓶
⓷
![Page 52: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/52.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.448.4 ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.nwater/nammonia = coeffwater/coeffammonia
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⓷
![Page 53: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/53.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.448.4 ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.nwater/nammonia = coeffwater/coeffammonia ➠ nwater = (nammonia)
(coeffwater)/coeffammonia
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⓷
![Page 54: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/54.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.448.4 ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.nwater/nammonia = coeffwater/coeffammonia ➠ nwater = (nammonia)
(coeffwater)/coeffammonia
➠ nwater = (48.4 mol)(6)/4
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⓷
![Page 55: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/55.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.448.4 ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.nwater/nammonia = coeffwater/coeffammonia ➠ nwater = (nammonia)
(coeffwater)/coeffammonia
➠ nwater = (48.4 mol)(6)/4 = 72.6 mol
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⓷
![Page 56: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/56.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.448.4 ?
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.nwater/nammonia = coeffwater/coeffammonia ➠ nwater = (nammonia)
(coeffwater)/coeffammonia
➠ nwater = (48.4 mol)(6)/4 = 72.6 mol
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⓷
![Page 57: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/57.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.448.4 72.6
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.nwater/nammonia = coeffwater/coeffammonia ➠ nwater = (nammonia)
(coeffwater)/coeffammonia
➠ nwater = (48.4 mol)(6)/4 = 72.6 mol
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⓷
![Page 58: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/58.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.448.4 72.6
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.nwater/nammonia = coeffwater/coeffammonia ➠ nwater = (nammonia)
(coeffwater)/coeffammonia
➠ nwater = (48.4 mol)(6)/4 = 72.6 mol
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⓷
![Page 59: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/59.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.448.4 72.6
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.nwater/nammonia = coeffwater/coeffammonia ➠ nwater = (nammonia)
(coeffwater)/coeffammonia
➠ nwater = (48.4 mol)(6)/4 = 72.6 mol
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⓷
![Page 60: Stoichiometry](https://reader037.fdocuments.us/reader037/viewer/2022110405/568132ae550346895d995fb7/html5/thumbnails/60.jpg)
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?b. How many moles of H2O are formed?
Ṃ (g/mol) 17.04 32.00 20.01 18.01
4 NH3(g) + 5 O2(g) →4 NO(g) + 6 H2O(g)
m (g) 824
n (mol) 48.448.4 72.6
1. Calculate the number of mols of NH3.nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol
2. Calculate the number of mols of NO.nNO/nammonia = coeffNO/coeffammonia ➠ nNO =
(nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol
3. Calculate the number of mols of H2O.nwater/nammonia = coeffwater/coeffammonia ➠ nwater = (nammonia)
(coeffwater)/coeffammonia
➠ nwater = (48.4 mol)(6)/4 = 72.6 mol