1. 2HCl + Ca(OH)2 → CaCl2 + 2H2O We don’t have enough calcium hydroxideb)
g HCl
mol HCl
mol Ca(OH)2 needed
g Ca(OH)2 needed
1 mol HCl
1 mol Ca(OH)2
g Ca(OH)2
1 mol Ca(OH)2 x
g HCl x x =
233.7g Ca(OH)2
needed
36.46 g HCl
2 mol HCl
Given: g Ca(OH)2
We don’t have enough
calcium hydroxide
Thus, Ca(OH)2 is the limiting reagent

2. c)
HCl d)
Use limiting reagent, Ca(OH)2, to calculate amount of HCl that
reacts with it.
g Ca(OH)2
mol Ca(OH)2
mol HCl reacted
g HCl reacted
36.46 g HCl
1 mol Ca(OH)2
g Ca(OH)2 x
2 mol HCl
1 mol Ca(OH)2 x
g Ca(OH)2 x =
1 mol HCl
g HCl reacted
Thus
starting with g HCl
g HCl reacted
23.32 g HCl in excess

3. Complete the other three reactions for next class
g Ca(OH)2
mol Ca(OH)2
mol CaCl2
g CaCl2
1 mol Ca(OH)2
g Ca(OH)2 x
1 mol CaCl2
1 mol Ca(OH)2 x
gCaCl2
1 mol CaCl2 x =
g Ca(OH)2
g CaCl2
g CaCl2
% Yield =
Actual Yield
Theoretical Yield
x 100 f)
294.62
% Yield =
x 100
93.7% =
314.55
Complete the other three reactions for next class