In the auto-reduction stage of copper extraction, copper(I) oxide (\text{Cu}_2\text{O}) re… — JAMB UTME

In the auto-reduction stage of copper extraction, copper(I) oxide (

\text{Cu}_2\text{O}

) reacts with copper(I) sulfide (

\text{Cu}_2\text{S}

) according to the following balanced equation:

\text{Cu}_2\text{S (s)} + 2\text{Cu}_2\text{O (s)} \rightarrow 6\text{Cu (s)} + \text{SO}_2\text{ (g)}

14.3\text{ g}

of copper(I) oxide reacts completely with excess copper(I) sulfide, what mass of metallic copper in grams is produced?

[\text{Cu} = 63.5, \text{O} = 16.0]

Answer: 19.05 g

Answer

The mass of metallic copper produced is 19.05 g.

According to the balanced chemical equation, 2 moles of copper(I) oxide react with copper(I) sulfide to yield 6 moles of metallic copper, which simplifies to a 1:3 molar ratio. Given that the molar mass of Cu2O is 143 g/mol, 14.3 g represents 0.1 mol of Cu2O. Based on the 1:3 ratio, this produces 0.3 mol of copper metal, corresponding to 19.05 g of Cu.

Step-by-Step Solution

Calculate the molar mass of copper(I) oxide

Molar mass of Cu2O = 143.0 g/mol

Required to convert the given mass of reactant to moles.

Determine moles of Cu2O reacted

Moles of Cu2O = 14.3 g / 143.0 g/mol = 0.10 mol

Establishes the quantitative amount of Cu2O in the chemical system.

Determine moles of copper metal produced using stoichiometry

Moles of Cu = 0.10 mol * (6 / 2) = 0.30 mol

The balanced chemical equation shows 2 moles of Cu2O yield 6 moles of Cu metal.

Calculate the mass of copper metal produced

Mass of Cu = 0.30 mol * 63.5 g/mol = 19.05 g

Converts the stoichiometric amount of product moles into grams.

Key Concept

Auto-reduction in copper extraction and stoichiometric mass calculations

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