Final answer:
To determine the volume of a 0.150 M HCl solution that can react with 8.40 g of CaCO₃, we use stoichiometry and the balanced chemical equation. The volume of the HCl solution is approximately 2.67 mL.
Step-by-step explanation:
To determine how many milliliters of a 0.150 M HCl solution can react with 8.40 g of CaCO₃, we need to use stoichiometry and the balanced chemical equation for the reaction. The balanced equation is:
2HCl(aq) + CaCO₃(s) → CaCl₂(aq) + CO₂(g) + H₂O(l)
From the equation, we can see that 2 moles of HCl react with 1 mole of CaCO₃. First, we need to calculate the number of moles of CaCO₃:
moles of CaCO₃ = mass / molar mass = 8.40 g / (40.08 g/mol + 12.01 g/mol + 3 * 16.00 g/mol) ≈ 0.2 mol
Since the mole ratio between HCl and CaCO₃ is 2:1, the number of moles of HCl needed is 2 times the number of moles of CaCO₃:
moles of HCl = 2 * 0.2 mol = 0.4 mol
Finally, we can calculate the volume of the HCl solution using the molarity:
volume = moles / molarity = 0.4 mol / 0.150 M ≈ 2.67 mL