Question

If 2.50 kg of copper ore (Cu3FeS3) is reacted with 0.60 kg of oxygen gas and the process has an 86.3% yield, what mass of pure copper is produced?

Answer 1

880.7 grams of pure copper is produced.

Step-by-step explanation:

Let's make the reaction:

2 Cu₃FeS₃ + 7O₂ = 6Cu + 2FeO + 6O₂

In this equation, 2 moles of Cu₃FeS₃ reacts with 7 moles of O₂

Mass / Molar mass = Moles

Molar mass Cu₃FeS₃ = 342,6 g/m

2500 g/342,6 g/m = 7.29 moles Cu₃FeS₃

600 g/ 32g/m = 18.75moles O₂

I have to find out my limiting reactant.

2 moles of Cu₃FeS₃ reacts with 7 moles of O₂

7.29 moles of Cu₃FeS₃ reacts with 25.51 moles

The O₂ is the limiting.

If 7 moles of O₂ produce 6 moles of Cu

18.75 moles of O₂ will produce, (18.75 . 6) / 7 = 16.07 moles

This is the 100 % yield, so:

100% _____ 16.07 moles of Cu produced

86.3% ______ (86.3 . 16.07) / 100 = 13.8 moles

Molar mass of Cu = 63.5 g/m

Moles . molar mass = mass

13.8 m . 63.5 g/m = 880.7 g