Question

30g piece of a marble was put into excess of

dil. HCl. When the reaction was completed

3360 cm3

of CO2 was obtained at STP. The

percentage of CaCO3 in the marble is?​

Answer 1

Answer:

50%

Explanation:

no. of moles of CO2 produced = 3360/ 22400 = 3/20

CaCO3 + 2HCl --> CaCl2 + 2H2O + CO2

so mass of CaCO3 in marble = 3 × 100/20 =15 gm

% of CaCO3 in marble = 50%

Answer 2

Answer:

The marble contains 50% of CaCO₃.

Explanation:

The CaCO₃ in the marble will react with dilute HCl in the following way

CaCO₃+2HCl→CaCl₂+H₂O+CO₂

100g of CaCO₃≡3360ml of CO₂

∴30g of CaCO₃≡$\frac{22400*30}{100}$=6720ml of CO₂

6720ml of CO₂ is produced when the marble contains 100% CaCO₃.

It is given that 3360ml of CaCO₃ is produced.

∴ The % percentage is given by

$\frac{100*3360}{6720}$=50%

Therefore, the marble contains 50% of CaCO₃.