A 62.0 mL sample of 1.0 M NaOH is mixed with 58.0 mL of 1.0 M H_2SO_4 in a large

Question

A 62.0 mL sample of 1.0 M NaOH is mixed with 58.0 mL of 1.0 M H_2SO_4 in a large Styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 25.2 degrees Celsius. After adding the NaOH solution to the coffee cup, the mixed solutions are stirred until the reaction is complete. Assume that the denisty  of the mixed solutins is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18J/(g*degrees Celsius), and that no heat is lost to the surrounding. The delta H rxn for the neautralization of NaOH with H_2SO_4 is -114kL/mol H_2SO_4. What is the maximum measured temperature in the Styrofoam cup?

Explanation / Answer

(62.0 mL) x (1.0 M NaOH) = 62.0 mmol NaOH

(58.0 mL) x (1.0 M H2SO4) = 58.0 mmol H2SO4

2 NaOH + H2SO4 ? Na2SO4 + 2 H2O

62.0 mmoles of NaOH would react completely with 62.0 x (1/2) = 31 mmoles of H2SO4, but there is more H2SO4 present than that, so H2SO4 is in excess and NaOH is the limiting reactant.

(0.031 mol H2SO4) x (-114 kJ/mol H2SO4) = 3534 J produced by the reaction and absorbed by the water

Supposing additive volumes:
(3534 J) / (4.18 J/(g

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