A copper ore consisting of 12.5% copper(II) sulfide 95.61 g/mol) , when heated and reacted with oxygen gas 32.00 g/mol) , produces copper(II) oxide (79.55 g/mol) and sulfur dioxide (64.06 g/mol) .If the reaction has a 90.0% yield, how many grams of copperII) oxide are produced when 1.00 kg of the copper ore is processed?
2 CuS(s) +3 O₂ (g) $\rightarrow$ 2 CuO(s) + 2 SO₂ (g)

Question

A copper ore consisting of 12.5% copper(II) sulfide 95.61 g/mol) , when heated and reacted with oxygen gas 32.00 g/mol) , produces copper(II) oxide (79.55 g/mol) and sulfur dioxide (64.06 g/mol) .If the reaction has a 90.0% yield, how many grams of copperII) oxide are produced when 1.00 kg of the copper ore is processed?
2 CuS(s) +3 O₂ (g)

\rightarrow

2 CuO(s) + 2 SO₂ (g)

Quizplus · Accepted Answer

First, calculate the mass of copper(II) sulfide (CuS) in 1.00 kg of the ore: $1.00 \, \text{kg} \times 12.5\% = 125 \, \text{g}$.The molar mass of CuS is 95.61 g/mol, so the moles of CuS in 125 g are $125 \, \text{g} / 95.61 \, \text{g/mol} = 1.307 \, \text{mol}$.The balanced chemical equation shows that 2 moles of CuS produce 2 moles of CuO. Therefore, 1.307 moles of CuS will produce 1.307 moles of CuO.The molar mass of CuO is 79.55 g/mol, so the mass of CuO produced is $1.307 \, \text{mol} \times 79.55 \, \text{g/mol} = 103.9 \, \text{g}$.Considering the 90.0% yield, the actual mass of CuO produced is $103.9 \, \text{g} \times 90.0\% = 93.51 \, \text{g}$, which rounds to 93.7 g.

A Copper Ore Consisting of 12 →\rightarrow→ 2 CuO(s) + 2 SO2 (G) A)93

A Copper Ore Consisting of 12 $\rightarrow$ 2 CuO(s) + 2 SO₂ (G)
A)93