Question 7 of 10 Map Sapling Learning In a calorimeter, 65.0 mL of0920 M H2SO4 w

Question

Question 7 of 10 Map Sapling Learning In a calorimeter, 65.0 mL of0920 M H2SO4 was added to 65.0 mL of 0.280 M NaOH. The reaction caused the temperature of the solution to rise from 21.29 C to 23.20 C. If the solution has the same density and specific heat as water 1.00 g/mL and 4.184 Jlg K, respectively), what is AH for this reaction (per mole of H20 produced)? individual volumes. Assume that the total volume is the sum of the Number kJ/ mol H,0 AH There is a hint available! View the hint by clicking on the divider bar again to hide the hint Previous 8 Give up & view solution check Answer Next Exit u Hint

Explanation / Answer

H2SO4 + 2 NaOH Na2SO4 + 2 H2O

(65.0 mL) x (0.920 M H2SO4) = 59.8 mmol H2SO4

(65.0 mL) x (0.280 M NaOH) = 18.2 mmol NaOH

18.2 mmoles of NaOH would react completely with 18.2 x (1/2) = 9.1 mmoles of H2SO4,
but there is more H2SO4 present than that, so H2SO4 is in excess and NaOH is the limiting reactant.

Number of moles of water formed = (18.2 mmol NaOH) x (2 mol H2O / 2 mol NaOH) = 18.2 mmol H2O = 0.0182 mol H2O

Heat released = m c T = (65.0 g + 65.0 g)x(4.184 J/g·K)x (23.20 - 21.29)°C = 1039 J gained by the solution

Hrxn = (1039 J) / (0.0182 mol H2O) = 57088 J

Expressed in kilojoules and rounded to three sig figs, the enthalpy change of reaction per mole of water produced will thus be:

Hrxn = - 57.1 kJ/mol H2O

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