what is the limiting reactant for Pb(NO3)2(aq)+2NaI(aq)--> PbI2(s)+2NaNO3(aq) when 78 g Pb(NO3)2 reacts with 32.5 g NaI?
answer : NaI
solution:
Pb(NO3)2(aq) + 2NaI(aq)------------------------> PbI2(s) + 2NaNO3(aq)
331.2 g 300g
78 g 32.5 g
331.2 g Pb(NO3)2 needs -----------------> 300 g NaI
78 g Pb(NO3)2 needs ----------------------> x g NaI
x = 300 x 78 / 331.2
x = 70.65 g
for 78g Pb(NO3)2 we need 70.65 g NaI . but we have only 32.5 g NaI . that means it is lesser amount than required .the product is based on NaI. so limiting reagent NaI
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