Everything Science Grade 11

6.8 CHAPTER 6. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE

m = n× M = 0. 1 × 87 .911 = 8. 79 g

The mass of iron (II) sulphide that is produced during this reaction is 8.79g.

Example 19: Industrial reaction to produce fertiliser

QUESTION

Sulphuric acid (H 2 SO 4 ) reacts with ammonia (NH 3 ) to produce the fertiliserammonium sulfate

((NH 4 ) 2 SO 4 ) according to the following equation:

H 2 SO 4 (aq) + 2NH 3 (g)→ (NH 4 ) 2 SO 4 (aq)

What is the maximummass of ammonium sulphate that can be obtained from 2.0 kg of

sulphuric acid and 1.0 kg of ammonia?

SOLUTION

Step 1 : Convert the mass of sulphuric acid and ammonia into moles

n(H 2 SO 4 ) =

m

M

=

2000 g

98. 078 g· mol−^1

= 20. 39 mol

n(NH 3 ) =

1000 g

17. 03 g· mol−^1

= 58. 72 mol

Step 2 : Use the balanced equation to determine whichof the reactants is limiting.

From the balanced chemical equation, 1 mole ofH 2 SO 4 reacts with 2 moles of

NH 3 to give 1 mole of (NH 4 ) 2 SO 4. Therefore 20.39 moles of H 2 SO 4 need to

react with 40.78 molesof NH 3. In this example, NH 3 is in excess and H 2 SO 4 is

the limiting reagent.

Step 3 : Calculate the maximumamount of ammonium sulphate that can be produced

Again from the equation, the mole ratio of H 2 SO 4 in the reactants to (NH 4 ) 2 SO 4

in the product is 1:1. Therefore, 20.39 moles ofH 2 SO 4 will produce 20.39 moles

of (NH 4 ) 2 SO 4.

The maximum mass of ammonium sulphate thatcan be produced is calcu-

lated as follows:

m = n× M = 20. 41 mol× 132 g· mol−^1 = 2694 g

The maximum amount of ammonium sulphate that can be produced is 2.694 kg.