Question

A quantity of ore consist of 90% (by weight) chalcopyrite mineral with a composition of CuFeS₂. The rest is inert quartz, SiO₂. The ore is roasted by heating it in oxygen at 800^oC. This process converts the ore into Fe₂O₃ and CuO. The two oxides are separated from each other and from the quartz using density differences and subsequently the oxides are reduced to the metal using carbon. A total of 323 kg of iron metal is obtained.

What is the mass of sulfur dioxide set free? _________ kg

What is the mass of carbon dioxide set free? _______kg

What is the mass of quartz tailings produced? ______ kg

Answers need to be within 1% accurate.
Molar masses: Fe: 55.8, Cu: 63.5, S: 32, O: 16, C: 12, Si: 28 [g/mol]

Answer 1

reaction is as per the equations shown

Given

Iron metal= 323kg

Moles of iron metal = 323/55.8 = 5.788 moles

4 moles iron is given by 2 moles Fe2O3

5.788 moles iron is given by 5.788*2/4 = 2.894moles of Fe2O3

(1)1 mole Fe2O3 gives 4 moles SO2 as by product

2.894 moles Fe2O3 gives =11.576 moles of SO2

=740.864 kg of SO2

(2) 1 mole Fe2O3 gives 2 moles Cuo

2.894 moles Fe2O3 gives 5.788 moles of CuO

2 moles Fe2O3 gives 3 moles CO2

2.894 moles give 4.341 moles of CO2

2 moles CuO gives 1mole CO2

5.788 moles give 2.894 moles CO2

Total Co2 = 7.208moles

= 7.208*44 = 317.152 kg

(3) 2 moles cufes2 gives 1 mole Fe2O3

2.894 moles moles is given by 5.788 moles of CuFeS2

Weight of CuFeS2 = 5.788*183.3= 1060.94 kg

This is 90% of the ore

Weight of the ore = 1060.9404*100/90

=1178.822 kg

Rest of the ore is quartz = 1178.822-1060.9404

=117.882 kg