Chemistry 2005 JAMB Past Questions - Question 26: How many moles of limestone will be required to produce 5.6g of CaO?

Correct Answer

D

Explanation

CaCO3 ------------- CaO +CO2therefore 5.6 = 0.10 mole
To determine the number of moles of limestone (CaCO₃) required to produce 5.6g of CaO, we need to consider the chemical reaction:

\[ CaCO₃ \rightarrow CaO + CO₂ \]

The molar mass of CaCO₃ is the sum of the atomic masses of calcium (Ca), carbon (C), and three oxygen (O) atoms:

\[ Molar \, mass \, of \, CaCO₃ = 40.08 \, g/mol + 12.01 \, g/mol + (3 \times 16.00 \, g/mol) \]

\[ Molar \, mass \, of \, CaCO₃ = 40.08 \, g/mol + 12.01 \, g/mol + 48.00 \, g/mol \]

\[ Molar \, mass \, of \, CaCO₃ = 100.09 \, g/mol \]

Now, we can use the molar mass to calculate the number of moles of CaCO₃ corresponding to 5.6g of CaO:

\[ \text{Number of moles of CaCO₃} = \frac{\text{Mass of CaO}}{\text{Molar mass of CaCO₃}} \]

\[ \text{Number of moles of CaCO₃} = \frac{5.6 \, g}{100.09 \, g/mol} \]

\[ \text{Number of moles of CaCO₃} \approx 0.0559 \, mol \]

Therefore, the correct answer is approximately \(0.056 \, mol\). None of the given options match exactly, but \(0.056 \, mol\) is closest to \(0.056 \, mol\) (considering significant figures), so the answer is closest to \(0.056 \, mol\).