To determine the limiting reagent in the reaction between impure calcium carbonate (CaCO₃) and dilute hydrochloric acid (HCl), we need to consider the stoichiometry of the balanced reaction:
\[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{CO}_2 (g) + \text{H}_2\text{O} (l) \]
From the equation, one mole of calcium carbonate reacts with two moles of hydrochloric acid.
Steps to determine the limiting reagent:
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Calculate moles of CaCO₃ (calcium carbonate):
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The molar mass of CaCO₃ is approximately:
- Ca = 40.08 g/mol
- C = 12.01 g/mol
- O = 16.00 g/mol × 3 = 48.00 g/mol
- Total = 40.08 + 12.01 + 48.00 ≈ 100.09 g/mol.
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Moles of CaCO₃ in the sample: \[ \text{Moles of CaCO₃} = \frac{12.5 , \text{g}}{100.09 , \text{g/mol}} \approx 0.125 , \text{mol} \]
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Determine how many moles of HCl are required:
- According to the balanced equation, 1 mole of CaCO₃ requires 2 moles of HCl. \[ \text{Moles of HCl required} = 2 \times \text{Moles of CaCO₃} \approx 2 \times 0.125 , \text{mol} = 0.250 , \text{mol} \]
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Determine the available moles of HCl:
- If we don't have a specific amount of HCl provided in the problem, we cannot calculate the exact limiting reagent without knowing how much HCl is present. However, if we assume excess HCl is available, then CaCO₃ would be the limiting reagent since it would be entirely used up.
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Conclusion:
- In typical scenarios where a limited amount of either reactant is available, the limiting reagent will be the one that runs out first. In this case, if enough dilute hydrochloric acid is available to fully react with the 0.125 mol of CaCO₃, then CaCO₃ (calcium carbonate) is the limiting reagent as it determines the amount of products formed.
Therefore, the limitng reagent in this reaction is CaCO₃ (calcium carbonate).