Answer:
1) Limiting reactant: Pb(C2H3O2)2
2) Theoretical Yield: 1,237 gr
3) Percent Yield: 80,27%
Explanation:
The limiting reactant is the reactant that is consumed first in the reaction. We find it by dividing the amount of moles of each reagent by its molar coefficient in the balanced reaction, the lowest value corresponds to the limiting reagent.
The theoretical yield is how much product is produced using the limiting reactant as the starting point.
The percent yield is the fraction between the real products obtained and the theoretical yield, all multiplied by 100.
K2SO4(aq)+Pb(C2H3O2)2(aq)→2KC2H3O2(aq)+PbSO4(s)
Limiting reactant:
The equation is already balanced, now we find the moles of each reactant and divide it by their molar coefficient:
K2SO4 (molar coefficient=1):
\(58 ml * \frac{1 L}{1000 mL}*\frac{0,122 moles}{1 L}=\frac{0,00708 moles}{1}=0,00708 moles\)
Pb(C2H3O2)2 (molar coefficient=1):
\(40 ml * \frac{1 L}{1000 mL}*\frac{0,102 moles}{1 L}=\frac{0,00408moles}{1}=0,00408 moles\)
As 0,00408 is lower than 0,00708, Pb(C2H3O2)2 is the limiting reactant.
Theoretical yield:
Now we find the amount of PbSO4 produced starting from the limiting reactant.
\(0,00408 moles*\frac{1 mol of PbSO4}{1 mol} * \frac{303,2gr}{1 mol PbSO4} =1,237 g PbSO4\)
Percent yield:
\(%Yield = \frac{0,993 }{1,237}*100=80,27%\)
