1. 2H2(g)+ O2(g)→ 2H2O(l) Thus, H2 is the limiting reagent
2 mol H2 =
2.44 mol H2
1.22 mol O2 x
1 mol O2
Given: 2.05 mol H2
2 mol H2O
g H2O
2.05 mol H2 x x =
36.93 g H2O
2 mol H2
1 mol H2O
Thus, H2 is the limiting reagent
36.93 H2O produced
synthesis

2. Mg(OH)2 is the limiting reactant
Mg(OH)2 + 2HCl → MgCl2 + 2H2O
1 mol Mg(OH)2
2 mol HCl
g HCl
5.87 g Mg(OH)2 x x x =
g Mg(OH)2
1 mol Mg(OH)2
1 mol HCl
Given: g HCl
HCl is in excess
7.33 g HCl
1 mol Mg(OH)2
1 mol MgCl2
g MgCl2
5.87 g Mg(OH)2 x x x =
g Mg(OH)2
1 mol Mg(OH)2
1 mol MgCl2
9.58 g MgCl2
Mg(OH)2 is the limiting reactant
0.10 mol MgCl2 formed
double replacement
1 mol MgCl2
9.58 g MgCl2 x
g MgCl2

3. Theoretical yield of H2O
HI + NaOH → NaI + H2O
1 mol NaOH
1 mol H2O
g H2O
18 g NaOH x x x =
g NaOH
1 mol NaOH
1 mol H2O
8.11 g H2O produced
Theoretical yield of H2O
% Yield =
Actual Yield
Theoretical Yield
x 100
7.9
% Yield =
x 100
97.4% =
8.11