Question

Ammonia is formed according to the reaction below. A chemist mixes 21 grams of nitrogen gas and 18 grams of hydrogen gas in a 2.0 L vessel. How many grams of hydrogen gas will be consumed?

Answer 1

Answer : The mass of hydrogen gas consumed will be, 4.5 grams

Explanation : Given,

Mass of $N_2$ = 21 g

Mass of $H_2$ = 18 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $H_2$ = 2 g/mole

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

$\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{21g}{28g/mole}=0.75moles$

$\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{18g}{2g/mole}=9moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

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

From the given balanced reaction, we conclude that

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

So, 0.75 moles of $N_2$ react with $3\times 0.75=2.25$ 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 because it limits the formation of product.

The moles of hydrogen gas consumed = 2.25 mole

Now we have to calculate the mass of hydrogen gas consumed.

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

$\text{Mass of }H_2=(2.25mole)\times (2g/mole)=4.5g$

Therefore, the mass of hydrogen gas consumed will be, 4.5 grams