Question

56 gm of N2(nitrogen) reacts with 30 gm of H2(hydrogen) to form NH3 ( ammonia). Calculate the no. of moles of NH3 formed.

Answer 1

Given :

➔ Nitrogen (56g) reacts with hydrogen (30g) to form ammonia.

To Find :

⟶ No. of moles of ammonia formed.

SoluTion :

➢ Stoichiometry deals with the calculation of moles of the reactants and the products involved in a chemical reaction.

➢ The substance which is completely consumed in the reaction and limits the amount of product formed is called limiting reagent.

✴ Balanced chemical reaction :

$\bigstar\:\underline{\boxed{\bf{\green{N_2+3H_2}\:\purple{\longrightarrow}\:\red{2NH_3}}}}$

⇒ 6g H$_2$ rects with 28g of N$_2$

Therefore, 30g H$_2$ reacts with

• 30/6 × 28 = 140g of N$_2$

✭ Hence, N$_2$ is the limiting reagent.

➠ 28g of N$_2$ produces 2 moles of NH$_3$

Therefore, 56g of N$_2$ will produces

• 56/28 × 2 = 4 moles of NH$_3$

Answer : 4 moles of NH$_3$