Question

When 28 g of nitrogen and 6 g of hydrogen react, 34 g of ammonia are produced. if 100 g of nitrogen react with 6 g of hydrogen, how much ammonia will be produced?

Answer 1

Answer : The mass of $NH_3$ produced will be, 34 grams.

Explanation : Given,

Mass of $N_2$ = 100 g

Mass of $H_2$ = 6 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 $H_2$.

$\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{100g}{28g/mole}=3.57moles$

$\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{6g}{2g/mole}=3moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction we conclude that

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

So, the given 3 moles of $N_2$ react with 1 moles of $H_2$

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 $NH_3$.

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

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

Now we have to calculate the mass of $NH_3$.

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

$\text{Mass of }NH_3=(2mole)\times (17g/mole)=34g$

Therefore, the mass of $NH_3$ produced will be, 34 grams.

Answer 2

N2 + 3H2 --> 2NH3
When 100g of N2 , no of moles of N2= 100/(28)=3.57 mol
no. of moles of h2 = 6/(2)=3mol
Therefore h2 is limiting reagent.
no. of moles of ammonia= 3/3*2=2moles
mass of ammonia produced= 2 mol * (14+3)= 34g