Question

What volume of carbon(IV)oxide is produced at s.t.p when 7.5g of calcium trioxocarbonate(IV) reacts with excess hydrochloric acid

Answer 1

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