CaCO3 --> CaO + CO2 20.7 g of CaCO3 reacts and 5.81 grams of CaO are produced. What is the percent yield of CaO?

Answer 1

1. First step is to identify what the percent yield is for a substance.

Percent yield= (Amount of products actually formed) / (Amount of products should form theoretically) x 100

2. The question gives the value for the numerator for the above equation, which is 5.81 g. Since it is given in mass units we are going to work on the rest of the problem in mass units.

3. Then, we need to find theoretical yield, which can be calculated from the amount of CaCO3 used. For that, we need to find how many moles of CaCO3 (molar mass is 100 g/mol) has been reacted to give CaO (molar mass is 56 g/mol).

Moles of CaCO3 used = 20.7 g/ (100 g/mol) = 0.207 mol

Since One CaCO3 gives one CaO moles,

Theoretical amount of CaO produced = 0.207 mol

Mass of theoretical CaO mass = 0.207 mol x 56 g/mol = 11.6 g

4.We have everything to calculate percent yield.

Percent yield = 5.81 g / 11.6 g x 100 = 50.1%