1.  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 + O 2  MgO

2. Limiting Reactant

3.  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.

4. 2 H 2 + O 2  2 H 2 O If you have 2 moles H 2 and 3 moles O 2, which is the limiting reactant? LR- H 2 You only need one mole of O 2 to react with the 2 moles of H 2 (you have an excess of 2 mol O 2 [3-1=2]). ER- O 2  2 mol H 2 X 1 mol O 2 = 1 mol O 2 2 mol H 2

5.  7.6 mol H 2 X 1 mol O 2 = 3.8 mol O 2 2 mol H 2 The 3.5 mol O 2 you have is smaller than the 3.8 mol O 2 that you would need to react with all the H 2 so the O 2 is the limiting reactant. 2 H 2 + O 2  2 H 2 O If you have 7.6 moles H 2 and 3.5 moles O 2, which is the limiting reactant? Needed to react with 7.6 mol H 2

6.  Remember to balance the equation first!

7.  2.0 mol HF X 1 mol SiO 2 = 0.50 mol SiO 2 4 mol HF The 4.5 moles SiO 2 you are given are MORE than enough to react with the 2.0 mol HF so HF is our limiting reactant SiO 2 + HF  SiF 4 + H 2 O If you have 2.0 mol HF and 4.5 mol SiO 2, which is the limiting reactant? 4 2

8. 15 g HCl X 1 mol HCl = 0.41 mol HCl 36.5g HCl HCl + Ca(OH) 2  CaCl 2 + H 2 O If you have 15 g HCl and 12 g Ca(OH) 2, which is the limiting reactant? 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.

9.  Zn + HCl  ZnCl 2 + H 2 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.

10.  Zn + HCl  ZnCl 2 + H 2 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 ZnCl 2 X 136.3 g ZnCl 2 = 132.57 g ZnCl 2 36.5g HCl 2 mol HCl 1 mol ZnCl 2

11.  If you did the reaction on the last two slides and you only collected 110 g of ZnCl 2, what is your % yield? % Yield = mass collected X 100% mass expected % Yield = 110 g X 100% = 82.97% 132.57 g

12.  % yield = experimental yield x 100% theoretical yield % expressing what you u collected vs what you were supposed to collect.