Question

10 g of caco3 are treated with a solution of hydrochloric acid.

Answer 1

Caco3 + 2HCl = Cacl2 +Co2 + H2O

1 mole of Caco3 = 100 g = 22.4 l

0.1 mole of Caco3 = 10g = 2.24 l or 2240 cm³

Answer 2

Answer:

CaCO3(s) + 2HCl(aq)→H2O(l) + CO2(g) + CaCl2(aq)CaCO3(s) + 2HCl(aq)→H2O(l) + CO2(g) + CaCl2(aq)  

No. of moles of CaCO3 (g) = Mass/Molar Mass = 10/100 = 0.1 moles

1 mole of CaCO3 (s) produces = 1 mole of CO2 gas

0.1 mole of CaCO3 (s) produces = 0.1 moles of CO2 (g)

1 mole at STP = 22.4 L

0.1 mole = 2.24 litre

Since only 1.12 litre of carbon dioxide is produced ,it means HCl is the limiting reagent.

No. of moles of CO2 in 1.12 litre = 1.12/22.4 moles =0.05 moles of CO2

1 mole of CO2 is produced by = 2 moles of HCl

​0.05 moles of CO2 is produced by = 2 x0.05 =0.1 mole of HCl

​

Thus No. of  moles of HCl present in the solution =0.1 moles