When heated, calcium carbonate decomposes according to the equation given below: CaCO3 --> CaO + CO2
If 50.0 g of CaCO3 react but only 20.0 g of CO2 are recovered, what is the percent yield of this reaction?

90.9%
40.0%
66.7%
250%

Percent yield = 90.9%

Explanation:

Given data:

Mass of CaCO₃ = 50.0 g

Mass of CO₂ produced = 20.0 g

Percent yield = ?

Solution:

Chemical equation:

CaCO₃ →  CaO + CO₂

Number of moles of CaCO₃:

Number of moles = mass/molar mass

Number of moles = 50.0 g/ 100.1 g/mol

Number of moles = 0.5 mol

Now we will compare the moles of CO₂ with CaCO₃.

                      CaCO₃          :        CO₂

                          1                :           1

                       0.5               :          0.5

Mass of CO₂: Theoretical yield

Mass = number of moles × molar mass

Mass = 0.5 mol × 44 g/mol

Mass = 22 g

Percent yield:

Percent yield = ( actual yield / theoretical yield ) × 100

Percent yield = (20.0 g/ 22.0 g) × 100

Percent yield = 0.909 × 100

Percent yield = 90.9%

- lead + phosphoric acid = hydrogen + lead(ii) phosphate