# Grams of N_2 reacts with 84.29 grams of H_2 in the chemical equation: N_2 + H_2-> NH_3 How do you balance the chemical equation and determine the limiting reactant? How much NH_3 is produced?

Jan 6, 2016

$\frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {H}_{2} \left(g\right) \rightarrow N {H}_{3} \left(g\right)$
If you have 84.29 g dihydrogen, there are $\frac{84.29 \cdot \cancel{g}}{2.02 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $41.7$ $m o l$ ${H}_{2}$. Stoichiometry requires 27.8 mol dinitrogen ($=$ ?? g). I think you have omitted some of the starting conditions in the problem