Sure! Let’s go step by step using the GRESA method to solve this problem.

GRESA Method

• Given:

  • Mass of CaCO₃ = 34.0 g
  • Molarity of HCl (HCl concentration) = 0.80 M
  • The balanced equation: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

• Required:

  • Volume of 0.80 M HCl needed to dissolve 34.0 grams of CaCO₃.

• Equation(s): First, we need to use the stoichiometry of the reaction: $\text{CaCO₃ (s)} + 2 \text{HCl (aq)} \rightarrow \text{CaCl₂ (aq)} + \text{H₂O (l)} + \text{CO₂ (g)}$ From the equation, 1 mole of CaCO₃ reacts with 2 moles of HCl.

  Now, let’s determine how many moles of CaCO₃ are in 34.0 grams using its molar mass.

  Molar mass of CaCO₃: $\text{CaCO₃} = 40.08 \, (\text{Ca}) + 12.01 \, (\text{C}) + 3 \times 16.00 \, (\text{O}) = 100.09 \, \text{g/mol}$

  So, the number of moles of CaCO₃ in 34.0 grams: $\text{Moles of CaCO₃} = \frac{\text{Mass of CaCO₃}}{\text{Molar mass of CaCO₃}} = \frac{34.0 \, \text{g}}{100.09 \, \text{g/mol}} = 0.3396 \, \text{mol}$

• Solution: According to the balanced equation, 1 mole of CaCO₃ reacts with 2 moles of HCl. So, for 0.3396 moles of CaCO₃, the required moles of HCl will be: $\text{Moles of HCl} = 2 \times 0.3396 = 0.6792 \, \text{mol}$

  Now, we use the formula for molarity to calculate the volume of 0.80 M HCl required: $M = \frac{n}{V} \quad \Rightarrow \quad V = \frac{n}{M}$ Where:

  • $M$ is the molarity (0.80 M),
  • $n$ is the number of moles of HCl (0.6792 mol).

  So, the volume $V$ of HCl needed is: $V = \frac{0.6792 \, \text{mol}}{0.80 \, \text{mol/L}} = 0.849 \, \text{L}$

  Converting this to milliliters: $V = 0.849 \, \text{L} \times 1000 = 849 \, \text{mL}$

• Answer: The volume of 0.80 M HCl required to dissolve 34.0 grams of CaCO₃ is 849 mL.

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Summary:

You need 849 mL of 0.80 M HCl to dissolve 34.0 grams of CaCO₃.

Would you like more details or have any further questions? Here are some additional questions that could help expand your understanding:

1. How can we determine the limiting reactant in this reaction?
2. What would happen if we used a higher concentration of HCl?
3. How could we calculate the amount of CO₂ produced in the reaction?
4. What would happen if the temperature of the reaction was increased?
5. Can the molarity of HCl be calculated using a titration method?

Tip: When solving stoichiometry problems, always make sure that the units cancel out properly, and double-check your balanced chemical equation!