If you have 3.2 g of magnesium to start with (and plenty of oxygen) how many grams of MgO will be produced in this reaction?

Mg + O2 MgO

Unit 8 - StoichiometryLIMITING REACTANT

Back to our Cake Analogy 1 cake mix + 3 eggs + 1 cup water 1 cake

If you have 2 mixes, 3 eggs and 5 cups of water, how many cakes can you make?

In this case we would call the eggs the limiting reactant. We can only make one cake because we don’t have enough eggs to match the other materials that are available. In this case we would make 1 cake and have 1 mix and 4 cups water left over.

Limiting reactant in a chemical Rxn

2 H2 + O2 2 H2O

If you have 2 moles H2 and 3 moles O2, which is the limiting reactant?

LR- H2

You only need one mole of O2 to react with the 2 moles of H2 (you have an excess of 2 mol O2 [3-1=2]).

ER- O2

2 mol H2 X 1 mol O2 = 1 mol O2

2 mol H2

Calculating Limiting reactantAlways in moles!

7.6 mol H2 X 1 mol O2 = 3.8 mol O2

2 mol H2

The 3.5 mol O2 you have is smaller than the 3.8 mol O2 that you would need to react with all the H2 so the O2 is the limiting reactant.

2 H2 + O2 2 H2O

If you have 7.6 moles H2 and 3.5 moles O2, which is the limiting reactant?

Needed to react with 7.6 mol H2

Remember to balance the equation first!

Calculating Limiting reactantAlways in moles!

2.0 mol HF X 1 mol SiO2 = 0.50 mol SiO2

4 mol HF

The 4.5 moles SiO2 you are given are MORE than enough to react with the 2.0 mol HF so HF is our limiting reactant

SiO2 + HF SiF4 + H2O

If you have 2.0 mol HF and 4.5 mol SiO2, which is the limiting reactant?

4 2

Calculating Limiting reactantAlways goes through moles!

15 g HCl X 1 mol HCl = 0.41 mol HCl

36.5g HCl

HCl + Ca(OH)2 CaCl2 + H2O

If you have 15 g HCl and 12 g Ca(OH)2, which is the limiting reactant?

2 2

12 g Ca(OH)2 X 1 mol Ca(OH)2 = 0.16 mol Ca(OH)2

74 g Ca(OH)2

X 1 mol Ca(OH)2 = 0.205 mol Ca(OH)2 2 mol HCl

The Ca(OH)2 is the limiting reactant because we do not have enough to react with all of the HCl we are given.

Using limiting reactant for stoichiometry

Zn + HCl ZnCl2 + H2

If you are given 70 g Zn and 71 g HCl, what will be the mass of the salt produced?

2

71 g HCl X 1 mol HCl = 1.95 mol HCl 36.5g HCl

70 g Zn X 1 mol Zn = 1.07 mol Zn

65.4 g Zn

X 2 mol HCl = 2.14 mol HCl 1 mol Zn

The HCl is the limiting reactant because we do not have enough HCl to react with all of the Zn we are given.

Using limiting reactant

Zn + HCl ZnCl2 + H2

If you are given 70 g Zn and 71 g HCl, what will be the mass of the salt produced?

HCl is our limiting reactant

2

71 g HCl X 1 mol HCl X 1 mol ZnCl2 X 136.3 g ZnCl2 = 132.57 g ZnCl2

36.5g HCl 2 mol HCl 1 mol ZnCl2132.57 g is the theoretical yield for this reaction.

% yield

% yield = experimental yield x 100%

theoretical yield

% expressing what you u collected vs what you were supposed to collect.

% Yield

If you did the reaction on the last two slides and you only collected 110 g of ZnCl2, what is your % yield?

% Yield = mass collected X 100% mass expected

% Yield = 110 g X 100% = 82.97% 132.57 g

Mg + HCl -- > MgCl2 + H2

If you have 25.4 g Mg and combine it with 50.0 mL of 1.25 M HCl, what mass of MgCl2 will you collect?

Do Now 9/22 – Do this then discuss questions from your reading on gasses

2

Zn + HCl ZnCl2 + H2

If you are given 70.0 g Zn and 71.0 g HCl, what will be the mass of the salt produced?

If you did the reaction and you only collected 110 g of ZnCl2, what is your % yield?

Do Now 9/24 – Do this then discuss questions from your reading on gasses

A sample of a compound contains 0.99 g Na, 1.365 g S and 1.021 g O. Determine its empirical formula.

ATP is an important molecule in living cells. A sample with a mass of 0.8138 g was analyzed and found to contain 0.1927 g C, 0.02590 g H, 0.1124 g N, and 0.1491 g P. The remainder was O. Determine the empirical formula of ATP. Its formula mass was found to be 507 g mol-1. Determine the molecular formula.

Do Now 9/25

A sample of a compound contains 0.99 g Na, 1.365 g S and 1.021 g O. Determine its empirical formula.

ATP is an important molecule in living cells. A sample with a mass of 0.8138 g was analyzed and found to contain 0.1927 g C, 0.02590 g H, 0.1124 g N, and 0.1491 g P. The remainder was O. Determine the empirical formula of ATP. Its formula mass was found to be 507 g mol-1. Determine the molecular formula.