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The theoretical yield of acetate is 2607 g. The actual yield of acetate is 1066.8 g. The percentage yield of acetate is 41%.
If 1 mole of vinegar contains 6.02 x 10^23 particles
x moles of vinegar contains 9.02 x 10^24 particles
x = 1 mole x 9.02 x 10^24 /6.02 x 10^23
x = 15 moles of vinegar
The reaction is as follows;
2HC2H3O2 + CaCO3 -----> Ca(C2H3O2)2 + H2O + CO2
Since 2 moles of vinegar reacts with 1 mole of carbonate
x moles of vinegar reacts with 16.5 moles of carbonate
x = 2 moles x 16.5 moles/ 1 mole
x = 33 moles of vinegar
We can see that the vinegar is the reactant in excess hence the carbonate is the limiting reactant.
Theoretical yield = 16.5 moles x 158 g/mol = 2607 g
Actual yield = 6.35 moles x 158 g/mol = 1066.8 g
Percent yield = 1066.8 g/2607 g × 100/1
= 41%
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Answer:
3.8 L
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