Question

solution is mixed with 40.0 mL of a 0.128 M lead(II) acetate solution and the following precipitation reaction occurs:
K2SO4(aq)+Pb(C2H3O2)2(aq)→2KC2H3O2(aq)+PbSO4(s)
The solid PbSO4 is collected, dried, and found to have a mass of 1.01 g .



Identify the limiting reactant.
Pb(C2H3O2)2
PbSO4
K2SO4
KC2H3O2


My class has just started learning about limiting reactants and I do not understand how to break down the equation to find it. Could someone please explain the steps to find it?

Thanks!

Answer 1

find the final moles of PbSO4(s)

compare with the initial moles of Pb(C2H3O2)2(aq)

was all of the acetate converted?

if yes , then the acetate is the limiting reactant

if no , then the K2SO4(aq) is the limiting reactant

Answer 2

There are at least two ways for limiting reagent (LR) problems. I do them the long way. However, it appears to me you haven't posted the entire problem. What you have posted could be a LR problem but usually there are numbers for BOTH reactants. Please advise.
K2SO4(aq)+Pb(C2H3O2)2(aq)→2KC2H3O2(aq)+PbSO4(s)