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Question

When 11.25 moles of lead nitrate are made to react with 2.5 moles of chromic sulphate, precipitation of lead sulphate takes place. Calculate the moles of Cr(NO3)3 formed in the final solution. Assume that lead sulphate is completely insoluble.
(Molar mass of lead - 207 g/mol)

A
7
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B
3
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C
5
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D
2
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Solution

The correct option is C 5
3Pb(NO3)2+Cr2(SO4)33PbSO4+2Cr(NO3)3
Initial moles of Pb(NO3)2=11.25 moles
Initial moles of Cr2(SO4)3=2.5 moles
Calculating limiting reagent:
3 moles of Pb(NO3)2 react with 1 mole of Cr2(SO4)3
11.25 moles of Pb(NO3)2will react with 13×11.25=3.75 moles of Cr2(SO4)3
Hence, the limiting reagent is chromic sulphate.
1 mole of Cr2(SO4)3 produces 2 moles of Cr(NO3)3
2.5 moles of Cr2(SO4)3 will produce 2×2.5=5 moles of Cr(NO3)3
So, moles of Cr(NO3)3 produced =2×2.5=5 moles

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