Question

N2 and H2 react with each other to produce ammonia according to the equation

N2(g) + 3H2(g) → 2NH3(g)

(i) Calculate the mass of ammonia produced if 2.0 × 103

g of N2 reacts with 1.0 × 103

g of H2.

(ii) Which compound is limiting reagent?

(iii) Find the mass and name the excess compound?​

Answer 1

Explanation:

the balanced chemical equation is as follows:

N2+3H2→2NH3

The molar masses of nitorgen, hydrogen and ammonia are 28g/mol,2g/mol and 17g/mol respectively.

28g of nitrogen react with 6g hydrogen to from 34g of ammonia.

2.00 X 10³ g nitrogen will be react with 6/28 X 2.00 X 10³ = 428.6 hydrogen.

However, 1.00 X 10³g of dihydrogen are present.

hence, nitrogen is the limiting reagent.

(¡) mass of ammonia produced = 34/28x2x10³= 2428.57g

(¡¡) Hydrogen is the excess reagent. Hence, it will remain unreacted.

(¡¡¡) mass of hydrogen left unreacted = 1.00 x 10³ - 428.6 = 571.4 g