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.

Answer 1

Answer:

10.85 g of water

Explanation:

First we write the balanced chemical equation

$3NO_{2} +H_{2}O -->2HNO_{3} +NO$

Then we calculate the number of moles of nitric acid produced

n(HNO3) = $\frac{mass}{molar mass} =\frac{75.9g}{63.02g/mol}=1.2044 mol$

According to the balanced equation, water needed in moles is always half the number of moles of HNO3 produced. So since we will produce 1.2044 mol of HNO3, we will need 0.6022 mol of water. Now to calculate what mass that is:

mass(water)=number of moles*molar mass=0.6022mol*18.02g/mol=10.85g