# Question #000a8

Oct 27, 2015

The percentage yield is indeed 45%

#### Explanation:

Reaction: $C a C {O}_{3}$ $\rightarrow$ $C a O$ + $C {O}_{2}$

The number of moles of $C a C {O}_{3}$ is the mass of 60g divided by the molar mass of Calcium Carbonate:

$n = \frac{m a s s}{m o l a r m a s s}$

$n = \frac{60 g}{100.0869 \frac{g}{m o l}}$

n = 0.599479052 moles $C a C {O}_{3}$

Therefore theoretical yield of $C a O$ is:

moles x molar mass of $C a O$ = g of $C a O$
(one to one reaction therefore moles of $C a C {O}_{3}$ will be the same as moles of $C a O$)

0.599479052 moles x 56.0744g/mol = g of $C a O$
Theoretical yield of $C a O$ = 33.6154 g

Thus percentage yield is as follows:

% Yield = (actual mass/theoretical mass) x 100
% Yield = $\frac{15 g}{33.6154 g}$ x 100
% Yield = 44.62%

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

So therefore if 15g of $C a O$ is produced from heating 60g of $C a C {O}_{3}$ then the percentage yield is 45%.

Hope I helped :)