You are given a crushed sample that is a mixture of limestone (caco3), lime (cao), and sand. the calcium carbonate, limestone, is the only material present in the material that will decompose when heated. you subject a 6.4734 g sample of the mixture to strong heating and after the sample reaches a constant mass (no more mass is lost with additional heating), the sample has a final weight of 4.3385 g. what is the percentage of calcium carbonate present in the original mixture? (f. wt. caco3 = 100.1)

Question

Bauer

Chemistry

5 September, 04:13

You are given a crushed sample that is a mixture of limestone (caco3), lime (cao), and sand. the calcium carbonate, limestone, is the only material present in the material that will decompose when heated. you subject a 6.4734 g sample of the mixture to strong heating and after the sample reaches a constant mass (no more mass is lost with additional heating), the sample has a final weight of 4.3385 g. what is the percentage of calcium carbonate present in the original mixture? (f. wt. caco3 = 100.1)

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Skye · Answer 1

Answer is: percentage of calcium carbonate is 75%.

Chemical reaction: CaCO₃ → CaO + CO₂.

m₁ (sample) = 6,4734 g.

m₂ (sample) = 4,3385 g.

m (CO₂) = m₁ (sample) - m₂ (sample).

m (CO₂) = 6,4734 g - 4,3385 g.

m (CO₂) = 2,1349 g.

n (CO₂) = m (CO₂) : M (CO₂).

n (CO₂) = 2,1349 g : 44 g/mol.

n (CO₂) = 0,0485 mol.

From chemical reaction: n (CO₂) : n (CaCO₃) = 1 : 1.

n (CaCO₃) = 0,0485 mol.

m (CaCO₃) = 0,0485 mol · 100,1 g/mol.

m (CaCO₃) = 4,854 g.

percentage of calcium carbonate = 4,854 g : 6,4734 g · 100%.

percentage of calcium carbonate = 75%.