A 90.2 mL sample of 1.00 M NaOH is mixed with 45.1 mL of 1.00 M H2SO4 in a large

Question

A 90.2 mL sample of 1.00 M NaOH is mixed with 45.1 mL of 1.00 M H2SO4 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 21.05 °C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is 32.10 °C. Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18 J/(g·°C), and that no heat is lost to the surroundings.

Write a balanced chemical equation for the reaction that takes place in the Styrofoam cup. Remember to include phases in the balanced chemical equation.

Explanation / Answer

mass of solution = 90.2 + 45.1 = 135.3 g

specific heat = 4.18 J/ g oC

temperature rise = 32.01 - 21.05 = 11.05

Q = m Cp dT

   = 135.3 x 4.18 x 11.05

Q = 6249.37 J

moles of NaOH = 90.2 x 1 / 1000 = 0.0902

moles of H2SO4 = 45.1 x 1 / 1000 = 0.0451

balanced chemical reaction :

H2SO4 (aq) + 2 NaOH (aq) -------------> Na2SO4 (aq) + 2 H2O (l)

   1              2

0.0451     0.0902

delta H = - Q / n

            = - 6.25 / 0.0451

            = - 138.6 kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.