Question #000a8

1 Answer
Oct 27, 2015

Answer:

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 :)