Question

What is the mass in grams of H₂ that can be formed from 54.6 grams of NH₃ in the following reaction?

Answer 1

Answer : The mass of $H_2$ is, 9.64 grams.

Explanation : Given,

Mass of $NH_3$ = 54.6 g

Molar mass of $NH_3$ = 17 g/mol

Molar mass of $H_2$ = 2 g/mol

First we have to calculate the moles of $NH_3$.

$\text{Moles of }NH_3=\frac{\text{Given mass }NH_3}{\text{Molar mass }NH_3}$

$\text{Moles of }NH_3=\frac{54.6g}{17g/mol}=3.21mol$

Now we have to calculate the moles of $H_2$

The balanced chemical equation is:

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

From the balanced reaction we conclude that

As, 2 mole of $NH_3$ react to give 3 moles of $H_2$

So, 3.21 mole of $NH_3$ react to give $\frac{3}{2}\times 3.21=4.82$ mole of $H_2$

Now we have to calculate the mass of $H_2$

$\text{ Mass of }H_2=\text{ Moles of }H_2\times \text{ Molar mass of }H_2$

$\text{ Mass of }H_2=(4.82moles)\times (2g/mole)=9.64g$

Therefore, the mass of $H_2$ is, 9.64 grams.