Question

Balance the chemical equation given below and calculate the number of molecules of SO2 gas liberated on consuming three molecules of oxygen gas at stp

Cu2S +O2 ---) Cu2O + SO2

Answer 1

2Cu2S +3O2 ---) 2Cu2O + 2SO2

Answer 2

$Cu_2S + 2O_2$ → $2CuO + SO_2$

Explanation:

• This is an oxidation-reduction type reaction.

Copper(I) Sulphide + Dioxygen → Copper(II) oxide + Sulfur dioxide

$Cu_2S + O_2$ → $CuO + SO_2$

Reactants used are -

• $Cu_2S$ - Copper(I) Sulphide

• $O_2$  - Dioxygen

Products used are -

• $CuO$ - Copper(II) oxide

• $SO_2$ - Sulfur dioxide

To balance the equation we need to add a coefficient of 2 in front of oxygen and a coefficient of 2 in front of copper(II) sulfide.

∴ The balanced equation is as follows -

$Cu_2S + 2O_2$ → $2CuO + SO_2$