A method for producing pure copper metal involves the reaction of copper(I)sulfide with oxygen gas to give copper metal and sulfur dioxide.Suppose the yield of this reaction is 87%.What mass of a copper ore consisting of 46% copper(I)sulfide must be mined in order to produce 1.0 × 10³ kg (1.0 metric ton)of copper metal?

Question

Quizplus · Accepted Answer

To produce 1.0 × 10^3 kg of copper with an 87% yield, you need more copper(I) sulfide than the theoretical amount. The reaction is Cu2S + O2 → 2Cu + SO2. The molar mass of Cu2S is 159.16 g/mol, and Cu is 63.55 g/mol. For 1.0 × 10^3 kg of Cu, you need 1.0 × 10^3 kg / (2 * 63.55 g/mol) * 159.16 g/mol = 1.25 × 10^3 kg of Cu2S. Considering the 87% yield, you need 1.25 × 10^3 kg / 0.87 = 1.44 × 10^3 kg of Cu2S. Since the ore is 46% Cu2S, you need 1.44 × 10^3 kg / 0.46 = 3.13 × 10^3 kg of ore, which rounds to 3.2 × 10^3 kg.

A Method for Producing Pure Copper Metal Involves the Reaction