Answer
0.500 mole of $HNO_3$ and 0.250 mole of $NO$.
Work Step by Step
- As we have determined in 7.98a, this is the balanced chemical equation for the reaction:
$3NO_2(g) + H_2O(g) --\gt 2HNO_3(g) + NO(g)$
1. Find the mole-mole conversion factors:
According to the coefficients: 1 mole $H_2O$ = 2 moles $HNO_3$
$\frac{ 1 \space mole \space H_2O}{ 2 \space moles \space HNO_3}$ and $\frac{ 2 \space moles \space HNO_3}{ 1 \space mole \space H_2O}$
According to the coefficients: 1 mole $H_2O$ = 1 mole $NO$
$\frac{ 1 \space mole \space H_2O}{ 1 \space mole \space NO}$ and $\frac{ 1 \space mole \space NO}{ 1 \space mole \space H_2O}$
2. Use the mole-mole conversion factors to calculate the amount of equivalent $HNO_3$ in moles:
$ 0.250$ mole $H_2O \times \frac{ 2 \space moles \space HNO_3}{ 1 \space mole \space H_2O} = 0.500$ mole $HNO_3$
$ 0.250$ moles $H_2O \times \frac{ 1 \space mole \space NO}{ 1 \space mole \space H_2O} = 0.250$ mole $NO$
