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:
![\text{Moles of }CaCO_3=\frac{80g}{100g/mol}=0.8mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DCaCO_3%3D%5Cfrac%7B80g%7D%7B100g%2Fmol%7D%3D0.8mol)
For the given chemical reaction:
![CaCO_3\rightarrow CaO+CO_2](https://tex.z-dn.net/?f=CaCO_3%5Crightarrow%20CaO%2BCO_2)
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:
![\text{Mass of CaO}=(0.8mol\times 56g/mol)=44.8g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20CaO%7D%3D%280.8mol%5Ctimes%2056g%2Fmol%29%3D44.8g)
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:
![\% \text{yield}=\frac{39 g}{44.8g}\times 100\\\\\% \text{yield}=87.05\%](https://tex.z-dn.net/?f=%5C%25%20%5Ctext%7Byield%7D%3D%5Cfrac%7B39%20g%7D%7B44.8g%7D%5Ctimes%20100%5C%5C%5C%5C%5C%25%20%5Ctext%7Byield%7D%3D87.05%5C%25)
Hence, the % yield of the product is 87.05 %