# How many moles of N_2 reacted if 0.75 mol of NH_3 is produced?

$\frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {H}_{2} \left(g\right) \rightarrow N {H}_{3} \left(g\right)$; $0.38 \cdot m o l$ ${N}_{2}$ reacted.
Given the stoichiometrically balanced equation, if $0.75 \cdot m o l$ ammonia were produced, then at least $\frac{0.75 \cdot m o l}{2}$ dinitrogen reacted. Is this clear?