Question

Question #000a8

Answer 1

The percentage yield is indeed 45%

Explanation:

Reaction: `CaCO_3` `rarr` `CaO` + `CO_2`

The number of moles of `CaCO_3` is the mass of 60g divided by the molar mass of Calcium Carbonate:

`n=(mass)/(molar mass)`

`n = (60g)/(100.0869g/(mol))`

n = 0.599479052 moles `CaCO_3`

Therefore theoretical yield of `CaO` is:

moles x molar mass of `CaO` = g of `CaO`
(one to one reaction therefore moles of `CaCO_3` will be the same as moles of `CaO`)

0.599479052 moles x 56.0744g/mol = g of `CaO`
Theoretical yield of `CaO` = 33.6154 g

Thus percentage yield is as follows:

% Yield = (actual mass/theoretical mass) x 100
% Yield = `(15g)/(33.6154g)` x 100
% Yield = 44.62%

Which then can be rounded off to 45% if no decimal places are required.

So therefore if 15g of `CaO` is produced from heating 60g of `CaCO_3` then the percentage yield is 45%.

Hope I helped :)