Answer to Question #70977 in General Chemistry for Jamar Ham

Question

What volume of 0.1248 M HCl should be added to a 5.0 g sample of CaCO3 for the CaCO3 to react completely.

CaCO3 (s) + 2 HCl ==>> CaCL2 + H2O + CO2 (g)

Answer 1

As one can see from the reaction equation, two moles of HCl reacts with one mole of calcium carbonate:
n(CaCO_3 )=n(HCl)/2.
The number of the moles of HCl equals:

n(HCl)=c(HCl)·V(HCl)=0.1248 M·V(HCl).
Thus, the volume of HCl is:

V(HCl)=n(HCl)/(0.1248 M)= 2n(CaCO_3 )/(0.1248 M).

The number of the moles of calcium carbonate is its mass divided by its molar mass:

n(CaCO_3 )=m(CaCO_3 )/M(CaCO_3 ) = (5.0 g)/(100.09 g/mol)=0.0500 mol.

Finally, the volume of HCl solution:

V(HCl)= 2n(CaCO_3 )/(0.1248 M)= (2·0.0500 (mol))/(0.1248 M)=0.8006 (L).

Answer: 0.8006 (L)