Question

Consider the following balanced reaction. How many grams of water are required to form 75.9 g of HNO3? Assume that there is excess NO2 present. The molar masses are as follows: H2O = 18.02 g/mol, HNO3 = 63.02 g/mol.

3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
Answer
26.5 g H2O
10.9 g H2O
43.4 g H2O
21.7 g H2O
38.0 g H2O

Answer 1

Moles of HNO3 formed: 75.9g/63.02g/mol= 1.204 moles

Based on the reaction equation, 1 mole of water makes 2 moles of HNO3
So moles of water needed= 1.204/2= 0.602 moles

0.602moles*18.02g/mol= 10.9 grams

Answer: 10.9 g H2O