Question

2g of impure CaCO3 on heating produces 0.84 g of CaO. The percentage purity of CaCO3 sample is

Answer 1

Answer:

The percentage purity of 2g impure CaCO3 is 75%.

Explanation:

Step 1 : write down the formula for decomposition of Calcium Carbonate.

CaCO3 — > CaO + CO2

The mole ratio of CaCO3 to CaO is 1 : 1

Step 2 : Calculate the moles of CaO produced.

Molar mass of CaO = 40 + 16 = 56 g/mol

Moles of CaO = 0.84/56 = 0.015 moles.

Step 3 : Get the mass of CaCO3.

Moles of CaCO3 = 0.015 since the mole ratio is 1 : 1.

Molar mass of CaCO3 = 40 + 12 + 16 × 3 = 100 g/mol

Mass of CaCO3 = 100 × 0.015 = 1.5 grams.

Mass of Pure CaCO3 is 1.5 g.

Step 4 : Get the percentage purity.

1.5/2 × 100% = 75%

Answer 2

Explanation:

Caco3 decomposes to give cao and co2 if the masses of cao and co2 are ... 4.4g respectively by heating 12 g of an impure caco3 sample then the