How much heat is liberated at constant pressure if 0.515 g of calcium carbonate reacts with 32.7 mL of 0.498 M hydrochloric acid? CaCO₃(s)+ 2HCl(aq)→ CaCl₂(aq)+ H₂O(l)+ CO₂(g); Δ_rH° = -15.2 kJ/mol-rxn

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First, calculate the moles of CaCO3 and HCl. For CaCO3: 0.515 g / 100.09 g/mol (molar mass) = 0.00515 mol. For HCl: 32.7 mL * 0.498 M = 0.0163 mol. The reaction is limited by CaCO3 since it requires 2 moles of HCl per mole of CaCO3, and there's more than enough HCl. The heat liberated is then: 0.00515 mol * (-15.2 kJ/mol) = -0.0783 kJ, which rounds to -0.0782 kJ, matching option A.

How Much Heat Is Liberated at Constant Pressure If 0