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 ...
Question 1 Report
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.
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