Answer
$0.604$ g of $H_2$
Work Step by Step
1. Calculate the molar mass of $N_2$:
$N: 14.01g * 2= 28.02g $
$ \frac{1 mole (N_2)}{ 28.02g (N_2)}$ and $ \frac{ 28.02g (N_2)}{1 mole (N_2)}$
2. The balanced reaction is:
$N_2 + 3H_2 --\gt 2NH_3$
According to the coefficients, the ratio of $N_2$ to $H_2$ is 1 to 3:
$ \frac{ 3 moles(H_2)}{ 1 mole (N_2)}$ and $ \frac{ 1 mole (N_2)}{ 3 moles(H_2)}$
3. Calculate the molar mass for $H_2$:
$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)}$
4. Use the conversion factors to find the mass of $H_2$
$2.80g(N_2) \times \frac{1 mole(N_2)}{ 28.02g( N_2)} \times \frac{ 3 moles(H_2)}{ 1 mole (N_2)} \times \frac{ 2.016 g (H_2)}{ 1 mole (H_2)} = 0.604g (H_2)$