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Question

35 mL of 1.5 M Pb(NO3)2 solution is mixed with 15 mL of 0.2 M KI solution. The maximum number of moles of PbI2 that can be precipitated is

A
5.25×103
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B
3×103
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C
1.5×103
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D
2.625×103
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Solution

The correct option is C 1.5×103
Milli moles of Pb(NO3)2=35×1.5=52.5
Milli moles of KI=15×0.2=3.0Pb(NO3)2+2KIPbI2+2KNO3

2 moles of KI react with 1 mole of Pb(NO3)2
moles of Pb(NO3)2 react with 3 mmoles of KI = 1×3×1032=1.5×103
. Hence the limiting reagent is KI.
Moles of PbI2 formed =32×103=1.5×103

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