Chapter 12: Chemical Quantities Section 12.2: Using Moles.
-
Upload
imogen-briggs -
Category
Documents
-
view
224 -
download
0
Transcript of Chapter 12: Chemical Quantities Section 12.2: Using Moles.
![Page 1: Chapter 12: Chemical Quantities Section 12.2: Using Moles.](https://reader036.fdocuments.us/reader036/viewer/2022062300/56649d365503460f94a0df6d/html5/thumbnails/1.jpg)
Chapter 12: Chemical Quantities
Section 12.2: Using Moles
![Page 2: Chapter 12: Chemical Quantities Section 12.2: Using Moles.](https://reader036.fdocuments.us/reader036/viewer/2022062300/56649d365503460f94a0df6d/html5/thumbnails/2.jpg)
• Used to relate moles of one substance to moles of another substance
• Use coefficients in the equation to convert to
moles of other reactants or products
Balanced chemical equations
![Page 3: Chapter 12: Chemical Quantities Section 12.2: Using Moles.](https://reader036.fdocuments.us/reader036/viewer/2022062300/56649d365503460f94a0df6d/html5/thumbnails/3.jpg)
The following recipe serves 4:4 potatoes (.3 lb each)2 onions (.2 lb each)8 carrots (.1 lb each)4 stalks of celery (.05 lb each)1.4 lbs of water
What is the total mass of the soup?How would you make enough for 8?How about 1240?
**With a balanced chemical equation and number of moles, we can predict the exact amount of
reactant and product in a reaction**
![Page 4: Chapter 12: Chemical Quantities Section 12.2: Using Moles.](https://reader036.fdocuments.us/reader036/viewer/2022062300/56649d365503460f94a0df6d/html5/thumbnails/4.jpg)
There are 4 steps to follow:1) Write the balanced chemical equation2) Convert the given mass or volume to moles3) Use the coefficients in the chemical equation to set up a mole ratio (the coefficients are the # of moles!)
HINT: The substance you are solving for goes on TOP
4) Convert these moles back to mass or volume as required
![Page 5: Chapter 12: Chemical Quantities Section 12.2: Using Moles.](https://reader036.fdocuments.us/reader036/viewer/2022062300/56649d365503460f94a0df6d/html5/thumbnails/5.jpg)
Use:
Grams A ↔ Moles A ↔ Moles B ↔ Grams B molar mass coefficients molar mass
Steps
![Page 6: Chapter 12: Chemical Quantities Section 12.2: Using Moles.](https://reader036.fdocuments.us/reader036/viewer/2022062300/56649d365503460f94a0df6d/html5/thumbnails/6.jpg)
N2 + 3H2 2NH3
•This reaction can be stated as: 1 mole of nitrogen will react with 3 moles of hydrogen to produce 2 moles of ammonia.
![Page 7: Chapter 12: Chemical Quantities Section 12.2: Using Moles.](https://reader036.fdocuments.us/reader036/viewer/2022062300/56649d365503460f94a0df6d/html5/thumbnails/7.jpg)
#10) What mass of CO2 forms when 95.6 g of C3H8 burns?
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
3 C = 36.033 g C= 12.011g 8 H= 8.064 g 2 O = 32 gC3H8 = 44.097 g/mol CO2 = 44.011
g/mol
95.6 g C3H8 x 1 mol C3H8 x 3 mol CO2 x 44.011 g CO2 =
44.097g C3H8 1 mol C3H8 1 mol CO2
= 286.2 g CO2
Practice Problems (pg. 415)
![Page 8: Chapter 12: Chemical Quantities Section 12.2: Using Moles.](https://reader036.fdocuments.us/reader036/viewer/2022062300/56649d365503460f94a0df6d/html5/thumbnails/8.jpg)
#11) How many grams of fluorine are required to produce 10 g XeF6?
Xe (g) + 3F2 (g) → Xe F6 (s)
F2: 2 x 18.998 g = 37.996 g
Xe F6: 131.29 + (6 x 18.998) = 245.278 g
10 g XeF6 x 1 mol XeF6 x 3 mol F2 x 37.996 g F2 = 245.278 g XeF6 1 mol XeF6 1 mol F2
= 4.65 g F2
Practice (cont)
![Page 9: Chapter 12: Chemical Quantities Section 12.2: Using Moles.](https://reader036.fdocuments.us/reader036/viewer/2022062300/56649d365503460f94a0df6d/html5/thumbnails/9.jpg)
![Page 10: Chapter 12: Chemical Quantities Section 12.2: Using Moles.](https://reader036.fdocuments.us/reader036/viewer/2022062300/56649d365503460f94a0df6d/html5/thumbnails/10.jpg)
The reactant that runs out first in a reaction/ stops the reaction
LIMITING REACTANT
![Page 11: Chapter 12: Chemical Quantities Section 12.2: Using Moles.](https://reader036.fdocuments.us/reader036/viewer/2022062300/56649d365503460f94a0df6d/html5/thumbnails/11.jpg)
#1) What is the limiting reactant in producing water (H2O), 5 g H2 or 5 g of O2? (convert g reactant → mol of product)
2H2 + O2 → 2H2O
5 g H2 x 1 mol H2 x 2 mol H2O = 2.48 mol H2O
2.0158 g H2 2 mol H2 5 g O2 x 1 mol O2 x 2 mol H2O = 0.31 mol H2O
31.998 g O2 1mol O2
Limiting reactant is O2 (runs out first)
Practice Problems
![Page 12: Chapter 12: Chemical Quantities Section 12.2: Using Moles.](https://reader036.fdocuments.us/reader036/viewer/2022062300/56649d365503460f94a0df6d/html5/thumbnails/12.jpg)
2) Which is the limiting reactant in producing NH3, 3.75 g N2 or 3.75 g H2?
N2 + 3H2 → 2 NH3
3.75g H2 x 1 mol H2 x 2 mol NH3 = 1.24 mol NH3
2.0158 g H2 3 mol H2 3.75g N2 x 1 mol N2 x 2 mol NH3 = 0.27 mol NH3
28 g N2 1 mol N2
Limiting reactant is N2
Practice (cont)