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In a constant-pressure calorimeter, 75.0 mL of 0.830 M H2SO4 was added to 75.0 mL of 0.310 M NaOH. The reaction caused the temp...Asked by wisa bedalie
In a constant-pressure calorimeter, 50.0 mL of 0.300 M Ba(OH)2 was added to 50.0 mL of 0.600 M HCl. The reaction caused the temperature of the solution to rise from 21.73 °C to 25.82 °C. If the solution has the same density and specific heat as water, what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes............... i know that you have to multiply the mass of H2O by the specific heat and by delta T, and then divide by the moles of H2O, but i don't understand how to find the moles of H2O..
Answers
Cj
Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O
The reactants are present in equimolar amounts, so there is no excess or limiting reactants.
(0.0500 L) x (0.600 mol/L HCl) x (2 mol H2O / 2 mol HCl) = 0.0300 mol H2O
(4.184 J/g·°C) x (50.0 g + 50.0 g) x (25.82 - 21.73)°C = 1711.256 J
(1711.256 J) / (0.0300 mol H2O) = 57042 J/mol = 57.0 kJ/mol H2O
The reactants are present in equimolar amounts, so there is no excess or limiting reactants.
(0.0500 L) x (0.600 mol/L HCl) x (2 mol H2O / 2 mol HCl) = 0.0300 mol H2O
(4.184 J/g·°C) x (50.0 g + 50.0 g) x (25.82 - 21.73)°C = 1711.256 J
(1711.256 J) / (0.0300 mol H2O) = 57042 J/mol = 57.0 kJ/mol H2O
Anonymous
Make sure answer is negative