# 1.09 g H_2 is allowed to react with 10.2 g N_2, producing 1.12 g NH_3. What is the theoretical yield in grams for this reaction under the given conditions?

Mar 7, 2017

Less than 20%..............

#### Explanation:

We need a stoichiometric equation:

$\frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {H}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s N {H}_{3} \left(g\right)$

$\text{Moles of dinitrogen}$ $=$ $\frac{10.2 \cdot g}{28.02 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.364 \cdot m o l$.

$\text{Moles of dihydrogen}$ $=$ $\frac{1.09 \cdot g}{2.02 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.540 \cdot m o l$.

$\text{Moles of ammonia}$ $=$ $\frac{1.12 \cdot g}{17.03 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.0658 \cdot m o l$.

Clearly dihydrogen is the limiting reagent, and at most we can make $\frac{2}{3}$ equiv ammonia, i.e.

$\text{% yield}$ $\equiv$ (0.0658*mol)/(2/3xx0.540*mol)xx100%=??%