What is the maximum volume, in dm3, of CO2(g) produced when 1.00 g of CaCO3(s) reacts with 20.0 cm3 of 2.00 moldm–3 HCl(aq)?CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)Molar volume of gas = 22.7 dm3mol–1; Mr(CaCO3) = 100.00A. 12×20.0×2.01000×22.7B. 20.0×2.01000×22.7C. 1.0100.00×22.7D.
Question
What is the maximum volume, in dm3, of CO2(g) produced when 1.00 g of CaCO3(s) reacts with 20.0 cm3 of 2.00 moldm–3 HCl(aq)?CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)Molar volume of gas = 22.7 dm3mol–1; Mr(CaCO3) = 100.00A. 12×20.0×2.01000×22.7B. 20.0×2.01000×22.7C. 1.0100.00×22.7D.
Solution
To solve this problem, we first need to determine the number of moles of each reactant.
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Calculate the number of moles of CaCO3: The molar mass of CaCO3 is given as 100.00 g/mol. So, 1.00 g of CaCO3 is 1.00/100.00 = 0.01 mol.
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Calculate the number of moles of HCl: The concentration of HCl is given as 2.00 mol/dm3 and the volume is 20.0 cm3. To convert the volume from cm3 to dm3, divide by 1000. So, 20.0 cm3 = 20.0/1000 = 0.02 dm3. Therefore, the number of moles of HCl is 2.00 mol/dm3 * 0.02 dm3 = 0.04 mol.
From the balanced chemical equation, we can see that 1 mol of CaCO3 reacts with 2 mol of HCl to produce 1 mol of CO2. Therefore, the limiting reactant is CaCO3 because we have less than 0.02 mol of it.
- Calculate the volume of CO2 produced: The molar volume of CO2 is given as 22.7 dm3/mol. So, the volume of CO2 produced is 0.01 mol * 22.7 dm3/mol = 0.227 dm3.
Therefore, the maximum volume of CO2 produced is 0.227 dm3. The correct answer is not listed among the options provided.
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