Question

56 g of nitrogen reacts with 10 g of hydrogen to produce ammonia. Mass of ammonia formed is
(1) 68 g
(2) 34.8 g
(3) 56.7 g
(4) 66 g​

Answer 1

Answer : The mass of ammonia formed is, 56.7 grams.

Solution : Given,

Mass of $N_2$ = 56 g

Mass of $H_2$ = 10 g  

Molar mass of $N_2$ = 28 g/mole

Molar mass of $H_2$ = 2 g/mole

Molar mass of $NH_3$ = 17 g/mole

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

$\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{56g}{28g/mole}=2moles$

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{10g}{2g/mole}=5moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction we conclude that

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

So, 5 moles of $H_2$ react with $\frac{5}{3}=1.66$ moles of $N_2$

From this we conclude that, $N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2$ 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, 3 mole of $H_2$ react to give 2 mole of $NH_3$

So, 5 moles of $H_2$ react to give $\frac{5}{3}\times 2=3.333$ moles of $NH_3$

Now we have to calculate the mass of $NH_3$

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

$\text{ Mass of }NH_3=(3.333moles)\times (17g/mole)=56.7g$

Therefore, the mass of ammonia formed is 56.7 grams.