What volume of carbon (lV) oxide is produced at s.t.p. when 2.5g of CaCO3 reacts with excess acid according to the following equation? CaCO3 + 2HCI → CaCI2 ...

What volume of carbon (lV) oxide is produced at s.t.p. when 2.5g of CaCO₃ reacts with excess acid according to the following equation? CaCO₃ + 2HCI → CaCI2 +H₂O + CO₂ [CaCO₃ = 100; molar volume of a gas at s.t.p = 22.4dm³]

Answer Details

The balanced equation for the reaction between CaCO₃ and 2HCI is: CaCO₃ + 2HCI → CaCI₂ + H₂O + CO₂ The molar mass of CaCO₃ is 100 g/mol, which means that 100 g of CaCO₃ will produce 1 mole of CO₂. Therefore, 2.5 g of CaCO₃ will produce: 1 mole CO₂ = 22.4 dm³ at s.t.p. x moles CO₂ = (2.5 g / 100 g/mol) = 0.025 moles x = 0.56 dm³ (rounded to two decimal places) Therefore, the volume of CO₂ produced at s.t.p is 0.56 dm³. Hence, the answer is (d) 0.56 dm³.