The question is incomplete, the complete question is:
When heat is applied to 80 grams of CaCO3, it yields 39 grams of CaO Determine the percentage of the yield.
CaCO3→CaO + CO2
<u>Answer:</u> The % yield of the product is 87.05 %
<u>Explanation:</u>
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
......(1)
We are given:
Given mass of
= 80 g
Molar mass of
= 100 g/mol
Putting values in equation 1, we get:

For the given chemical reaction:

By stoichiometry of the reaction:
If 1 mole of
produces 1 mole of CaO
So, 0.8 moles of
will produce =
of CaO
We know, molar mass of
= 56 g/mol
Putting values in above equation, we get:

The percent yield of a reaction is calculated by using an equation:
......(2)
Given values:
Actual value of the product = 39 g
Theoretical value of the product = 44.8 g
Plugging values in equation 2:

Hence, the % yield of the product is 87.05 %