1.) Calculation: If 9.02 x 1024 particles of vinegar (HC2H3O2)HC2H3O2) are added to 16.5 moles of eggshell (CaCO3) and 6.35 mole

Question

3 years ago

10

s g of
calcium acetate is formed. What is the;
(a) theoretical yield,
(b) actual yield and,
(c) percent yield

1 answer:

blsea [12.9K] · Answer 1 · 2022-10-01T11:29:41+03:00

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%