In a constant-pressure calorimeter, 70.0 mL of 0.310 M Ba(OH)2 was added to 70.0 mL of 0.620 M HCl. The...?

Question

reaction caused the temperature of the solution to rise from 23.45 °C to 27.67 °C. what is ΔH for this reaction (KJ released per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes and that the density and heat capacity are equal to that of water.

Sp.h of water: 4.184 j /g c

Sp.h of water: 4.184 j /g c

Roger the Mole · Accepted Answer

(4.184 J/g·°C) x (70.0 g + 70.0 g) x (27.67 °C - 23.45 °C) = 2471.9 J produced

Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O

The reactants are in their stoichiometric ratio, so there is no excess and either could be considered the limiting reactant.
(0.0700 L) x (0.620 mol/L HCl) x (2 mol H2O / 2 mol HCl) = 0.0434 mol H2O

(2471.9 J) / (0.0434 mol H2O) = 56956 J/mol = 57.0 kJ/mol H2O

In a constant-pressure calorimeter, 70.0 mL of 0.310 M Ba(OH)2 was added to 70.0 mL of 0.620 M HCl. The...?

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