chem問題

Question

A mixture of zinc sulphide and zinc oxide is heated. After strong heating for 30 mins, 0.52g of a gaseous product A is formed. The residue is then heated strongly with excess carbon. 0.7g of another gaseous product B is formed after the reaction is completed.
a) Identity the two gaseous products in the reactions involved
b)Write balanced equations for the reactions as described above
c)calculate the percentage by mass of zinc oxide in the mixture

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Answer 1

(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%