Question

CaCO3(s) ∆→CaO(s) + CO2(g).

Answer 1

Answer:

74.9%.

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:

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

Number of moles of CaCO₃ available:

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

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:

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

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

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

What's the percentage yield of this reaction?

$\displaystyle \textbf{Percentage}\text{ Yield} = \frac{\textbf{Actual}\text{ Yield}}{\textbf{Theoretical}\text{ Yield}}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} = \frac{13.2}{17.632}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} =74.9\%$.