Question

The unbalanced equation for the reduction of copper(I) sulfide with oxygen is given below. What mass of copper is produced from the reduction of 8.80 g Cu2S

Answer 1

Answer: The mass of copper produced is 7.12 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

Given mass of copper (I) sulfide = 8.80 g

Molar mass of copper (I) sulfide = 156.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of copper (I) sulfide}=\frac{8.80g}{156.2g/mol}=0.056mol$

The chemical equation for the reaction of copper (I) sulfide and oxygen gas follows:

$Cu_2S+O_2\rightarrow 2Cu+SO_2$

By Stoichiometry of the reaction:

1 mole of copper (I) sulfide produces 2 moles of copper

So, 0.056 moles of copper (I) sulfide will produce = $\frac{2}{1}\times 0.056=0.112mol$ of copper

Now, calculating the mass of copper from equation 1, we get:

Molar mass of copper = 63.55 g/mol

Moles of copper = 0.112 moles

Putting values in equation 1, we get:

$0.112mol=\frac{\text{Mass of copper}{63.55g/mol}\\\\\text{Mass of copper}=(0.112mol\times 63.55g/mol)=7.12g$

Hence, the mass of copper produced is 7.12 grams