Question

What is the maximum mass in grams of NH3 that can be produced by the reaction of of 2.5 g N2 with 2.5 g of H2 via the equation below?
N2 (g) + 3 H2 (g) → 2 NH3 (g)

Answer 1

Answer: The mass of $NH_3$ produced is, 3.03 grams.

Explanation : Given,

Mass of $N_2$ = 2.5 g

Mass of $H_2$ = 2.5 g

Molar mass of $N_2$ = 28 g/mol

Molar mass of $H_2$ = 2 g/mol

First we have to calculate the moles of $N_2$ and $H_2$.

$\text{Moles of }N_2=\frac{\text{Given mass }N_2}{\text{Molar mass }N_2}=\frac{2.5g}{28g/mol}=0.089mol$

and,

$\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}=\frac{2.5g}{2g/mol}=1.25mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 1 mole of $N_2$ react with 3 mole of $H_2$

So, 0.089 moles of $N_2$ react with $0.089\times 3=0.267$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $N_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 1 mole of $N_2$ react to give 2 mole of $NH_3$

So, 0.089 mole of $N_2$ react to give $0.089\times 2=0.178$ mole of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

Molar mass of $NH_3$ = 17 g/mole

$\text{ Mass of }NH_3=(0.178moles)\times (17g/mole)=3.03g$

Therefore, the mass of $NH_3$ produced is, 3.03 grams.