Question

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 1

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