Question

A sample of impure sulphide ore contains 42.34% Zn. The percentage of ZnS in the sample is

Answer 1

Explanation:

ANSWER

Consider the volume V=100g hence Cu=66.6g

The number of moles =

63.6

66.6

=1.05 mole

1 mole of oxygen is per 2 moles Cu.

Therefore moles of oxygen =

2

1.05

=0.53 moles

The weight of oxygen is then 0.53 ml ×16 g/mol =8.42g

The weight of Cu

2

O= weight of Cu + weight of O

=66.6g+8.42g

=75.02g in 100g of sample

Therefore percentage of Cu

2

O in 100g of sample =

100

75.02

×100%=75%