Question

10 g impure sample of CaCO3(s) requires 4.38 g HCl for complete neutralisation. The percentage purity of the sample is

Answer 1

Answer:

Explanation:

- CaCO3 Δ CaO + CO2 Let pure sample of CaCO3 = x grams Mass of CaO produced after ... The percentage purity of calcium carbonate in the sample is: A . 60%. B . ... y=40 grams ... HCl and the solution was boiled. ... 12.3 ml of same hypo solution required 24.6 ml of 0.5 iodine for complete neutralization.

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Answer 2

Answer:

option a 60%

Explanation:

caso3 + 2hcl = cacl2 + h2o + co2

i.e., 2 mole of hcl needs 1 mole of caco3

so, 2(36.5)g of hcl needs 100g of caco3

1g of hcl needs 100/73 g of caco3

4.38 of hcl needs 100/73 x 4.38g of caco3 = 6g of caco3

% of purity of sample = 6/10 x 100% = 60%

( impure sample of caco3 = 10g ) given in question