Question

Calculate the percentage purity of CaCO3 when 50g of marble reacts with an excess of HCl to make 15 g of CO2.

Answer 1

$CaCO_3+2HCl→CaCl_2+H_2O+CO_2 ​$

From the reaction above, 1 mole of CO₂ is produced by 1 mole of CaCO₃.

This means that, 44g of CO₂ is produced by 100g of CaCO₃.

15g of CO₂ is therefore produced by $\frac{15×100}{44}$g of CaCO₃.

34g of CaCO₃ produces 15g of CO₂.

Percentage of purity = $\dfrac{\textsf{Mass of pure sample}}{\textsf{Total mass of impure sample}} ×100\%= \dfrac{34}{50}×100\%=68\%$

Therefore, the percentage purity of the marble is 68%.