What mass of caco3 is required to react completely with 0.56 l of hcl?

The balanced equation for the above reaction is as follows;
CaCO₃ + 2HCl > CaCl₂ + H₂O + CO₂
stoichiometry of CaCO₃ to HCl is 1:2
molar volume states that 1 mol of any gas occupies a volume of 22.4 L at STP.
volume of 22.4 L occupied by 1 mol
therefore 0.56 L occupied by - 0.56 L / 22.4 L/mol = 0.025 mol
number of HCl moles reacted - 0.025 mol
2 mol of HCl reacts with 1 mol of CaCO₃
therefore 0.025 mol reacts with - 0.025/2 = 0.0125 mol 
mass of CaCO₃ required - 0.0125 mol x 100 g/mol = 1.25 g 
1.25 g of CaCO₃ is required 

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