Question

CaCO3(s) ∆→CaO(s) + CO2(g).

If 13.2 g of CO2 was produced from the thermal decomposition of 40.1 g of CaCO3, what
is the percentage yield of the reaction?

Answer 1

74.9%.

Step-by-step explanation:

Relative atomic mass data from a modern periodic table:

• Ca: 40.078;
• C: 12.011;
• O: 15.999.

What's the theoretical yield of this reaction?

In other words, what's the mass of the CO₂ that should come out of heating 40.1 grams of CaCO₃?

Molar mass of CaCO₃:

`M(\text{CaCO}_3) = 40.078 + 12.011 + 3 * 15.999 = 100.086\;\text{g}\cdot\text{mol}^(-1)`.

Number of moles of CaCO₃ available:

`\displaystyle n(\text{CaCO}_3) =(m)/(M) = (40.1)/(100.086) = 0.400655\;\text{mol}`.

Look at the chemical equation. The coefficient in front of both CaCO₃ and CO₂ is one. Decomposing every mole of CaCO₃ should produce one mole of CO₂.

`n(\text{CO}2) = n(\text{CaCO}_3)= 0.400655\;\text{mol}`.

Molar mass of CO₂:

`M(\text{CO}_2) = 12.011 + 2* 15.999 = 44.009\;\text{g}\cdot\text{mol}^(-1)`.

Mass of the 0.400655 moles of
`\text{CO}_2` expected for the 40.1 grams of CaCO₃:

`m(\text{CO}_2) = n\cdot M = 0.400655 * 44.009 = 17.632\;\text{g}`.

What's the percentage yield of this reaction?

`\displaystyle \textbf{Percentage}\text{ Yield} = \frac{\textbf{Actual}\text{ Yield}}{\textbf{Theoretical}\text{ Yield}}* 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} = (13.2)/(17.632)* 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} =74.9\%`.