Answer
54.8 g of $HNO_3$.
Work Step by Step
1. Calculate the molar mass of $NO_2$:
Molar mass :
$N: 14.01g $
$O: 16.00g * 2= 32.00g $
14.01g + 32.00g = 46.01g
$ \frac{1 mole (NO_2)}{ 46.01g (NO_2)}$ and $ \frac{ 46.01g (NO_2)}{1 mole (NO_2)}$
2. The balanced reaction is:
$3NO_2(g) + H_2O(g) --\gt 2HNO_3(g) + NO(g)$
According to the coefficients, the ratio of $NO_2$ to $HNO_3$ is 3 to 2:
$ \frac{ 2 moles(HNO_3)}{ 3 moles (NO_2)}$ and $ \frac{ 3 moles (NO_2)}{ 2 moles(HNO_3)}$
3. Calculate the molar mass for $HNO_3$:
Molar mass :
$H: 1.008g $
$N: 14.01g $
$O: 16.00g * 3= 48.00g $
1.008g + 14.01g + 48.00g = 63.02g
$ \frac{1 mole (HNO_3)}{ 63.02g (HNO_3)}$ and $ \frac{ 63.02g (HNO_3)}{1 mole (HNO_3)}$
4. Use the conversion factors to find the mass of $HNO_3$
$60.0g(NO_2) \times \frac{1 mole(NO_2)}{ 46.01g( NO_2)} \times \frac{ 2 moles(HNO_3)}{ 3 moles (NO_2)} \times \frac{ 63.02 g (HNO_3)}{ 1 mole (HNO_3)} = 54.8g (HNO_3)$
