How many grams of Ca(NO₃)₂ can be produced by reacting excess HNO₃ with 7.40 g of Ca(OH)₂?

Question

Quizplus · Accepted Answer

The balanced chemical equation for the reaction is:
Ca(OH)2 + 2 HNO3 → Ca(NO3)2 + 2 H2O

First, we need to find the limiting reagent:
molar mass of Ca(OH)2 = 74.1 g/mol
7.40 g Ca(OH)2 x (1 mol Ca(OH)2 / 74.1 g Ca(OH)2) = 0.0998 mol Ca(OH)2

molar mass of HNO3 = 63.0 g/mol
0.0998 mol Ca(OH)2 x (2 mol HNO3 / 1 mol Ca(OH)2) x (1 L HNO3 / 16.7 mol HNO3) = 0.0119 L HNO3

The volume of HNO3 required to react with Ca(OH)2 is very small, so we can assume that HNO3 is in excess and Ca(OH)2 is limiting.

The balanced chemical equation for the reaction of Ca(OH)2 and HNO3 can be modified to give the equation for the formation of Ca(NO3)2:
Ca(OH)2 + 2 HNO3 → Ca(NO3)2 + 2 H2O

molar mass of Ca(NO3)2 = 164.1 g/mol
0.0998 mol Ca(OH)2 x (1 mol Ca(NO3)2 / 1 mol Ca(OH)2) x (164.1 g Ca(NO3)2 / 1 mol Ca(NO3)2) = 16.4 g Ca(NO3)2

Therefore, the answer is B) 16.4 g.

How Many Grams of Ca(NO3)2 Can Be Produced by Reacting

How Many Grams of Ca(NO₃)₂ Can Be Produced by Reacting