chem問題

(a)
A is sulphur dioxide, SO2
B is carbon dioxide, CO2

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(b)
When the mixture is strongly heated for 30 minutes, zinc sulphide reacts with oxygen in air to give zinc oxide and sulphur dioxide gas.
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
When the residue is strongly heated with carbon, zinc oxide is reduced to zinc and carbon is oxidized to carbon dioxide gas.
2ZnO(s) + C(s) → 2Zn(s) + CO2(g)

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(c)
Molar mass of SO2 = 32.1 + 16x2 = 64.1 g mol-1
Molar mass of ZnS = 65.4 + 32.1 = 97.5 g mol-1
Molar mass of ZnO = 65.4 + 16 = 81.4 g mol-1
Molar mass of CO2 = 12 + 16x2 = 44 g mol-1

Consider the reaction of heating the mixture in air.
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
Mole ratio ZnS : ZnO : SO2 = 1 : 1 : 1

Mass of SO2 formed = 0.52 g
No. of moles of SO2 formed = 0.52/64.1 = 0.00811 mol
No. of moles of ZnO formed = 0.00811 mol
No. of moles of ZnS in the mixture = 0.00811 mol
Mass of ZnS in the mixture = 0.00811 x 97.5 = 0.791 g

Consider the reaction of heating the residue with carbon.
2ZnO(s) + C(s) → 2Zn(s) + CO2(g)
Mole ratio ZnO : CO2 = 2 : 1

Mass of CO2 = 0.7 g
No. of moles of CO2 = 0.7/44 = 0.0159 mol
No. of moles of ZnO in the residue = 0.0159 x 2 = 0.0318 mol
No. of moles of ZnO formed from ZnS = 0.00811 mol
No. of moles of ZnO in the mixture = 0.0318 – 0.00811 = 0.0237 mol
Mass of ZnO in the mixture = 0.0237 x 81.4 = 1.93 g

The mixture contains 0.791 g of ZnS and 1.93 g of ZnO.
Percentage by mass of ZnO in the mixture
= (1.93 / (0.791+1.93)) x 100%
= 70.9%