Answer
142 g of $F_2$ are needed to react with 25.5 g of $NH_3$
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 $F_2$ is 2 to 5:
$ \frac{ 5 moles(F_2)}{ 2 moles (NH_3)}$ and $ \frac{ 2 moles (NH_3)}{ 5 moles(F_2)}$
3. Calculate the molar mass for $F_2$:
$F: 19.00g * 2= 38.00g $
$ \frac{1 mole (F_2)}{ 38.00g (F_2)}$ and $ \frac{ 38.00g (F_2)}{1 mole (F_2)}$
4. Use the conversion factors to find the mass of $F_2$
$25.5g(NH_3) \times \frac{1 mole(NH_3)}{ 17.03g( NH_3)} \times \frac{ 5 moles(F_2)}{ 2 moles (NH_3)} \times \frac{ 38.00 g (F_2)}{ 1 mole (F_2)} = 142g (F_2)$