Answer
$20.5$ g of $NH_3$
Work Step by Step
1. Calculate the molar mass of $H_2$:
$H: 1.008g * 2= 2.016g $
$ \frac{1 mole (H_2)}{ 2.016g (H_2)}$ and $ \frac{ 2.016g (H_2)}{1 mole (H_2)}$
2. The balanced reaction is:
$N_2 + 3H_2 --\gt 2NH_3$
According to the coefficients, the ratio of $H_2$ to $NH_3$ is 3 to 2:
$ \frac{ 2 moles(NH_3)}{ 3 moles (H_2)}$ and $ \frac{ 3 moles (H_2)}{ 2 moles(NH_3)}$
3. Calculate the molar mass for $NH_3$:
$N: 14.01g $
$H: 1.008g * 3= 3.024g $
14.01g + 3.024g = 17.03g
$ \frac{1 mole (NH_3)}{ 17.03g (NH_3)}$ and $ \frac{ 17.03g (NH_3)}{1 mole (NH_3)}$
4. Use the conversion factors to find the mass of $NH_3$
$3.64g(H_2) \times \frac{1 mole(H_2)}{ 2.016g( H_2)} \times \frac{ 2 moles(NH_3)}{ 3 moles (H_2)} \times \frac{ 17.03 g (NH_3)}{ 1 mole (NH_3)} = 20.5g (NH_3)$
