    Question

# Lead is formed from its ore by the following reactions: 2PbS(s)+3O2(g)→2PbO(s)+2SO2(g) (Yield = 40 %) PbO(s)+C(s)→Pb(s)+CO(g) (Yield = 80 %) If 550 kg of PbS was initially taken, find the amount of Pb in kg formed. (Molar mass of Pb=207 g/mol)

A
147.75 kg
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B
288.63 kg
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C
152.42 kg
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D
108.23 kg
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Solution

## The correct option is C 152.42 kg2PbS(s)+3O2→2PbO+2SO2 (i) PbO+C→Pb+CO (ii) Mol of PbS=550239×1000=2.301×1000=2301 moles As per the stoichiometry of the reaction, 2 mol of PbS produces 2 moles of PbO 2301 mol of PbS will form 2301 mol of PbO Yield of the reaction is given as 40 % Actual amount of PbO formed =2301×0.40=920.4 moles In reaction (ii), 1 mol of PbO produces 1 mol of Pb 920.4 mol will form 920.4 mol of Pb Yield of the reaction is given as 80 % Actual amount of Pb formed =920.4×0.8=736.32 moles Amount of Pb formed =736.32×207=152.42 kg  Suggest Corrections  0      Similar questions
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