a student reacted 0.500 g of Pb(NO3)2 with .750 g of KI according to the reactio

Question

a student reacted 0.500 g of Pb(NO3)2 with .750 g of KI according to the reaction Pb(NO3)2 + 2KI ---> PbI2 + 2KNO3 a) how many moles of Pb(NO3)2 were used? b) how many moles of KI were used? c) how many moles of PbI2 would form based on the moles of Pb(N)3)2 used? d) How many moles of PbI2 would form based on the moles of KI used? e) which is the limiting reactant? f) what is the theoretical yield of PbI2 in grams? g) if the student obtained .583 grams of PbI2 product after collecting it by filtration and drying it, what was the percent yield of PbI2 obtained?

Explanation / Answer

moles of Pb(NO3)2=0.5/331.2=1.51 x`10^-3

moles of KI=0.75/166=4.52 x10^-3

1 mole of Pb(No3)2 requires 2 moles of KI

So 1.51 x`10^-3 of Pb(No3)2 requires 3.02 x10^-3 moles of KI

a.)1.51 x`10^-3 of Pb(No3)2 are used

b.)3.02 x10^-3 moles of KI are used

c.)1.51 x`10^-3 of PbI2 formed

d.)1.51 x`10^-3 of PbI2 formed

e.)Pb(NO3)2

f.)PbI2 mass=1.51 x`10^-3*461=0.696 grams

g.)Percent yield=0.583*100/0.696=83.77%

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