Question

nitrogen and hydrogen react to form ammonia if 1000 g of H2 react with 2000 g of N2 .will any of the two reactants remain unreacted? if yes which one and what will be its mass? calculate the mass of ammonia that will be formed ​

Answer 1

N2(g) + H2(g) → 2NH3(g)

(i) Calculate the mass of ammonia produced if 2.00×103g dinitrogen reacts with 1.00×103g of dihydrogen.

(ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass?

Answer:-

1 mole of dinitrogen (28g) reacts with 3 mole of dihydrogen (6g) to give 2 mole of ammonia (34g).

∴ 2000 g of N2 will react with H2 = 6/28 ×200g = 428.6g. Thus, here N2 is the limiting reagent while H2 is in excess.

28g of N2 produce 34g of NH3.

∴2000g of N2 will produce = 34/28×2000g = 2428.57 g of NH3.

(ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 will remain unreacted.

(iii) Mass of dihydrogen left unreacted = 1000g – 428.6g = 571.4 g

Answer 2

Answer : The $H_2$ reactant remain unreacted and the mass of the unreacted $H_2$ reactant is, 428.58 grams. The mass of $NH_3$ produced will be, 2428.62 grams.

Explanation : Given,

Mass of $H_2$ = 1000 g

Mass of $N_2$ = 2000 g

Molar mass of $H_2$ = 2 g/mole

Molar mass of $N_2$ = 28 g/mole

Molar mass of $NH_3$ = 17 g/mole

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

$\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{1000g}{2g/mole}=500moles$

$\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{2000g}{28g/mole}=71.43moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$3H_2+N_2\rightarrow 2NH_3$

From the balanced reaction we conclude that

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

So, 71.43 moles of $N_2$ react with $\frac{3}{1}\times 71.43=214.29$ 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.  That means, the $H_2$ reactant remain unreacted.

The excess moles of $H_2$ reactant = 500 - 214.29 = 285.71 moles

The mass of excess $H_2$ reactant = $285.1mole\times 2g/mole=571.42g$

The mass of unreacted $H_2$ reactant = 1000 - 571.42 = 428.58 g

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

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

So, 71.43 moles of $N_2$ react to give $\frac{2}{1}\times 71.43=142.86$ moles of $NH_3$

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

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

$\text{Mass of }NH_3=(142.86mole)\times (17g/mole)=2428.62g$

Therefore, the mass of $NH_3$ produced will be, 2428.62 grams