Question

A sample of HCl has 20% hydrogen cloride.How many grams of this acidic sample is required for reacting completely with 50g of CaCO3.

Answer 1

Equation of the reaction between calcium carbonate and hydrochloric acid

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

The molar mass of CaCO3 is 100 g/mol. Thus the amount of 50g of CaCO3 is 0.5 mol.
One mol of calcium carbonate reacts with two mol of hydrochloric acid. Thus, 0.5 mol of CaCO3 reacts with 1 mol of HCl.
The molar mass of HCl is 36.5 g/mol. The mass of 1 mol HCl is 36.5 g.
The concentration of HCl in the solution by weight is
w(HCl) = m(HCl) / m(solution)
Thus, the mass of the 20% solution of 36.5 g of HCl is
m(HCl solution) = m(HCl) / w(HCl) = 36.5 g / 0.2 = 182.5 g.

Answer: (iv) 182.5 g
50g CaCO3 will react with 182.5 g of 20% HCl by weight.

Answer 2

CaCO3 (aq) + 2 HCl (aq) → CaCl​2 (aq)+ H2O(l) + CO2(g)



1 mole of Calcium Carbonate reacts with 2 mole of HCl.

 or

0.5 mole of Calcium Carbonate reacts with 1 mole of HCl.

Number of moles of Calcium Carbonate = 50100 = 0.5 mole.


The molar mass of HCl is 36.5 g/mol.

  20 g of HCl is present in 100 g of solution.


Thus, the mass of the 20% solution of 36.5 g of HCl is
m(HCl solution) = m(HCl)/w(HCl)

=36.5/ 0.2 = 182.5 g.