Question

Ammonia reacts with sulfuric acid to give the fertilizer ammonium sulphate. Calculate the volume of ammonia [at STP] used to form 59 g of ammonium sulphate.
[N=14, H=1, S=32, O=16]
A- 20.02 L
B-24.33 L
C-66.20 L
D-34.12 L

Answer 1

First of all, write the balanced chemical equation for the reaction of ammonia and sulfuric acid then apply volume of compound as per mole concept (i.e. 22.4 L volume for one mole of compound).

Complete Solution :
Firstly, write the balanced chemical equation for the given reaction.
$\text{Ammonia + Sulphuric acid }\to \text{ Ammonium sulphate}$
i.e. $\mathbf{2N}{{\mathbf{H}}_{3}}+{{\mathbf{H}}_{2}}\mathbf{S}{{\mathbf{O}}_{4}}\to {{\left( \mathbf{N}{{\mathbf{H}}_{4}} \right)}_{_{2}}}\mathbf{S}{{\mathbf{O}}_{4}}$
At S.T.P. i.e. Standard temperature and pressure condition volume of one mole of gas is 22.4 L.
-Volume of one mole of ammonia = 22.4 L
-Volume of two moles of ammonia is calculated as,
= $22.4\text{ x 2}$
= $44.8\,\text{L}$
(As 2 moles of ammonia are required for formation of one mole ammonium sulphate.)
- Molecular weight of ammonium sulphate is calculated as,
= $\left( \text{14 x 2} \right)\text{ + }\left( \text{1 x 8} \right)\text{ + }\left( \text{32 x 1} \right)\text{ + }\left( \text{16 x 4} \right)$
= 132 g
Volume of ammonia required to form 132g ammonium sulphate = 44.8 L
Volume of Ammonia required to form 59 g ammonium sulphate is calculated as,
= $\dfrac{44.8}{132} \times 59$
= 20.02 L
From the above statement we can conclude that the volume of ammonia required used to form 59g of ammonium sulphate is 20.02 L.
So, the correct answer is “Option A”.

Note: Beware of counting of the moles of compounds for calculation of volume of ammonia in the balanced equation. Convert the gas moles to volume at STP condition only as given in question too.