Answer: the limiting reactant is the lead(II) acetate solution
The theoretical yield is 1.092 g PbSO4
The percent yield is 91,3 %
Explanation:
We verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest result is the limiting reactant.
K2SO4(aq)+Pb(C2H3O2)2(aq) → 2KC2H3O2(aq)+PbSO4(s)
we multiply the giving volumen and the concentration, to find out the moles available for reacting and then we use the stoichiometric relation to find out the amount of the PbSO4 generated
[tex]0.061L x \frac{0.112 moles}{1L} x \frac{1mol PbSO4}{1mol K2SO4} x \frac{303.6g PbSO4}{1mol PbSO4} =2.06g PbSO4[/tex]
[tex]0.035L x \frac{0.104moles}{1L} x \frac{1mol PbSO4}{1 mol. lead(II) acetate} x\frac{303.26g PbSO4}{1 mol PbSO4} = 1.092 g PbSO4[/tex]
By comparing , the between 2.06g and 1.092g , the lower one is the limiting reactant.
So the lead(II) acetate solution is the limiting reactant. and after knowning this, we continue calculating the theoretical yield, and the percent yield.
The theorical yield is the amount of product that will be produced when all the limiting reactant is consumed (100%); the answer is already calculated above : 1.092g PbSO4 will be produced theorically.
Now the percent yield is the actual product / theorical product x 100
Percent yield = 0.997g / 1.092g X100
Percent yield = 91.3 %
The percent yield of PbSO4 is 90.6 %.
The equation of the reaction is;
K2SO4(aq) + Pb(C2H3O2)2(aq)----------> 2KC2H3O2(aq) + PbSO4(s)
Number of moles K2SO4 = 61.0/1000 L × 0.112M = 0.0068 moles
Number of moles of Pb(C2H3O2)2 = 35/1000 L × 0.104 = 0.0036 moles
Given that the reaction is 1:1, we can see that Pb(C2H3O2)2 is the limiting reactant.
The theoretical yield of PbSO4 is 0.0036 moles × 303.26 g/mol
= 1.1 g of PbSO4
The percent yield is obtained from;
actual yield/theoretical yield × 100/1
% yield = 0.997g/1.1 g × 100/1
% yield = 90.6 %
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