# 600 g of dinitrogen reacts with 100 g of dihydrogen to form ammonia. Find the limiting reagent and calculate the amount of ammonia formed? P.S- pls solve the question by calculating the moles

Dec 21, 2017

We assess the reaction: $\frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {H}_{2} \left(g\right) \rightarrow N {H}_{3} \left(g\right)$

#### Explanation:

$\text{Moles of dihydrogen} = \frac{100 \cdot g}{2.016 \cdot g \cdot m o {l}^{-} 1} = 49.6 \cdot m o l$

$\text{Moles of dinitrogen} = \frac{600 \cdot g}{28.02 \cdot g \cdot m o {l}^{-} 1} = 21.4 \cdot m o l$

Clearly, there is insufficient dihydrogen for complete reduction....and dihydrogen is the LIMITING reagent....

One third of an equivalent, i.e. $\frac{49.6 \cdot m o l}{3} = 16.5 \cdot m o l$ dinitrogen could react to give ammonia.... Note that we might expect incomplete reduction of dinitrogen in this scenario, i.e. hydrazine, ${H}_{2} N - N {H}_{2}$, could be quantitatively produced....