Answer
8.21 g $NO_{2}$.
Work Step by Step
Balanced chemical reaction:
3 $NO_{2}$ (g) + $H_{2}O$ ${\longrightarrow}$ 2 $HNO_{3}$ (aq) + NO (g)
The strategy to find the mass of $NO_{2}$ required to produce 7.50 g $HNO_{3}$ is:
g $HNO_{3}$ ${\rightarrow}$ mol $HNO_{3}$ ${\rightarrow}$ mol $NO_{2}$ ${\rightarrow}$ g$NO_{2}$
The conversion factors are:
$\frac{1 mol (HNO_{3})}{Molar mass (HNO_{3})}$ to convert to moles $HNO_{3}$, where Molar mass $HNO_{3}$:
Molar mass $HNO_{3}$ = 1 x 1.0 g + 1 x 14.0 g + 3 x 16.0 g = 63 g
$\frac{3 mol (NO_{2})}{2 mol (HNO_{3})}$, to convert the moles of $HNO_{3}$ to mol of $NO_{2}$.
$\frac{Molar mass(NO_{2})}{1 mol(NO_{2})}$ to convert to g of $NO_{2}$, where the molar mass of $NO_{2}$ is:
molar mass of $NO_{2}$ = 1 x 14 + 2 x 16 = 46 g
Now we use of the conversion factors to find the mass of NO2:
7.50 g $HNO_{3}$ x $\frac{1 mol (HNO_{3})}{Molar mass (HNO_{3})}$ x $\frac{3 mol (NO_{2})}{2 mol (HNO_{3})}$ x $\frac{Molar mass(NO_{2})}{1 mol(NO_{2})}$
7.50 g $HNO_{3}$ x $\frac{1 mol (HNO_{3})}{63 g (HNO_{3})}$ x $\frac{3 mol (NO_{2})}{2 mol (HNO_{3})}$ x $\frac{46g(NO_{2})}{1 mol(NO_{2})}$ = 8.214 g $NO_{2}$ or 8.21 g $NO_{2}$.
