Answer
$10.4$ g of $N_2F_4$ can be produced when 3.40 g of $NH_3$ reacts.
Work Step by Step
1. Calculate the molar mass of $NH_3$:
$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)}$
2. The balanced reaction is:
$2NH_3(g) + 5F_2(g) --\gt N_2F_4(g) + 6HF(g)$
According to the coefficients, the ratio of $NH_3$ to $N_2F_4$ is 2 to 1:
$ \frac{ 1 moles(N_2F_4)}{ 2 moles (NH_3)}$ and $ \frac{ 2 moles (NH_3)}{ 1 moles(N_2F_4)}$
3. Calculate the molar mass for $N_2F_4$:
$N: 14.01g * 2= 28.02g $
$F: 19.00g * 4= 76.00g $
28.02g + 76.00g = 104.02g
$ \frac{1 mole (N_2F_4)}{ 104.02g (N_2F_4)}$ and $ \frac{ 104.02g (N_2F_4)}{1 mole (N_2F_4)}$
4. Use the conversion factors to find the mass of $N_2F_4$
$3.40g(NH_3) \times \frac{1 mole(NH_3)}{ 17.03g( NH_3)} \times \frac{ 1 moles(N_2F_4)}{ 2 moles (NH_3)} \times \frac{ 104.02 g (N_2F_4)}{ 1 mole (N_2F_4)} = 10.4g (N_2F_4)$
