Question

Does anybody know how to do q4. Please show working out thanks.

Answer 1

Answer:

% purity of limestone = 96.53%

Explanation:

Question (4).

Weight of impure CaCO₃ = 25.9 g

Molecular weight of CaCO₃ = 40 + 12 + 3(16)

                                              = 100 g per mole

We know at S.T.P. number of moles of CO₂ = 1 and volume = 22.4 liters

From the given reaction, 1 mole of CaCO₃ reacts with 1 mole or 22.4 liters of

CO₂.

∵ 22.4 liters of CO₂ was produced from CaCO3 = 100 g

∴ 1 liter of CO₂ will be produced by CaCO₃ = $\frac{100}{22.4}$

∴ 5.6 liters of CO₂ will be produced by CaCO₃ = $\frac{100\times 5.6}{22.4}$

                                                                              = 25 g

Therefore, % purity of CaCO₃ = $\frac{\text{Weight calculated}}{{\text{Weight given}}}\times 100$

                                                 = $\frac{25}{25.9}\times 100$

                                                 = 96.53 %