CaCO3 + 2HCl = CaCl2 + H2O + CO2
n(CaCO3) = n(CO2) = "{\\frac {7.5}{40+12+48}}={\\frac {7.5}{100}}=0.075" mol.
At s.t.p. we can use The value of molar volume which is equal to 22.4 l/mol.
V(CO2) = 0.075*22.4 = 1.68 l
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