Question

12.5g of impure sample of limestone on heating gives 4.4g of CO₂ Tnhen percantage purity of CaCO₃ in the sample is

(1)72
(2)80
(3)85
(4)8

Answer 1

The percentage purity of CaCO3 in the limestone sample is calculated by comparing the mass of CO2 produced to the molar masses of CO2 and CaCO3, resulting in a purity of 80%.

Step-by-step explanation:

To calculate the percentage purity of CaCO3 in the limestone sample, we use the relationship between the mass of the carbon dioxide produced and the molar mass of calcium carbonate. The chemical equation for the decomposition of calcium carbonate is:

CaCO3 → CaO + CO2

From the equation, one mole of CaCO3 produces one mole of CO2. We know the molar mass of CaCO3 is 100 g/mol and the molar mass of CO2 is 44 g/mol. Given that 4.4 g of CO2 are produced, we have:

Mass of pure CaCO3 = (Mass of CO2 produced) * (Molar mass of CaCO3 / Molar mass of CO2)

Mass of pure CaCO3 = (4.4 g) * (100 g/mol / 44 g/mol) = 10 g

Now, we can calculate the percentage purity of CaCO3 in the sample:

Percentage purity = (Mass of pure CaCO3 / Mass of limestone sample) * 100%

Percentage purity = (10 g / 12.5 g) * 100% = 80%

Therefore, the percentage purity of CaCO3 in the limestone sample is 80%.