What volume of carbon (IV) oxide in dm 3 is produced at s.t.p. when 2.50g of CaCO3 reacts with excess acid according to the following equation? CaCO3(s) + 2HCI(aq) → CaCl2(aq) + H2O(1) + CO2(g)[CaCO2 100; molar volume of a gas at s.t.p. = 22.4dm 3]
Question
What volume of carbon (IV) oxide in dm 3 is produced at s.t.p. when 2.50g of CaCO3 reacts with excess acid according to the following equation? CaCO3(s) + 2HCI(aq) → CaCl2(aq) + H2O(1) + CO2(g)[CaCO2 100; molar volume of a gas at s.t.p. = 22.4dm 3]
Solution
Step 1: Identify the balanced chemical equation The balanced chemical equation is already given: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
Step 2: Calculate the number of moles of CaCO3 The molar mass of CaCO3 is given as 100 g/mol. So, the number of moles of CaCO3 is given by the formula:
Number of moles = mass / molar mass = 2.50g / 100 g/mol = 0.025 mol
Step 3: Use the stoichiometry of the reaction From the balanced chemical equation, we can see that 1 mole of CaCO3 produces 1 mole of CO2. Therefore, 0.025 moles of CaCO3 will produce 0.025 moles of CO2.
Step 4: Calculate the volume of CO2 at STP The molar volume of a gas at STP is given as 22.4 dm^3/mol. So, the volume of CO2 produced is given by the formula:
Volume = number of moles * molar volume = 0.025 mol * 22.4 dm^3/mol = 0.56 dm^3
So, the volume of carbon (IV) oxide (CO2) produced at STP when 2.50g of CaCO3 reacts with excess acid is 0.56 dm^3.
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