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 349 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]
Hint There is a similar problem on wikis.lib.ncsu.edu, see http://wikis.lib.ncsu.edu/index.php/CH_101/Exercises_3 under example 6.

Answer 1

349,000g Fe x [1 mole / 55.8g] = 6254.48029 moles Fe ==> 6254.48029 moles CuFeS2.

6254.48029 moles CuFeS2 ==> 12508.9606 moles S ===> 12508.9606 moles SO2 x [64gSO2/mole] = 800573.478 g SO2
800.57 kg SO2

6254.48029 moles Fe ==> 3127.24015 moles Fe2O3 ==> 9381.72045 moles O
6254.48029 moles Cu ==> 6254.48029 moles CuO ==> 6254.48029 moles O
Total moles O = 15636.2007 moles O ==> 7818.10035 moles CO2 x [44gCO2 / 1mole] = 343996.415 g CO2
344 kg CO2

6254.48029 moles CuFeS2 x [183.3gCuFeS2 / mole] = 1146446.24 g CuFeS2.
1146.45 kg CuFeS2 = 0.90 x total ore mass
Total ore mass = 1145.45 kg / 0.9 = 1273.83 kg ore

1256kg ore - 1130kgCuFeS2 = 126kg SiO2