# How much ammonia (NH3) can be obtained when 3.0g H2 reacts with 100.g N2?

${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) \rightarrow 2 N {H}_{3} \left(g\right)$
We have the stoichiometric equation above. Clearly, the limiting reagent is dihydrogen gas, as we have only $\frac{3.0 \cdot \cancel{g}}{2 \cdot \cancel{g} \cdot m o {l}^{- 1}}$ $=$ ?? $m o l$.