A 64.0 mL sample of 1.0 M NaOH is mixed with 44 0 mL of 1.0 M H_2SO_4 in a large

Question

A 64.0 mL sample of 1.0 M NaOH is mixed with 44 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 29.2 degree 0. After adding the NaOH solution to the coffee cup. the mixed solutions are stirred until reaction is complete 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 middot a^C), and that no heat is lost to the surroundings. the Delta H_r times n for the neutralization of NaOH with H_2SO_4 is - 114 kj/mole H_2SO_4 What is the maximum measured temperature in the Styrofoam cup? the heat of reaction is absorbed by the water in the cup and raises its temperature How do you calculate the temperature rise associated with the addition of heat?

Explanation / Answer

moles of NaOH = 64 x 1 / 1000 = 0.064 mol

moles of H2SO4 = 44 x 1 / 1000 = 0.044 mol

2 NaOH + H2SO4   ----------------> Na2SO4 + 2 H2O

    2                1

0.064           0.044

here limiting reagent is NaOH. so we consider this moles to solve the problem

delta H = - Q / n

- 114 = - Q / 0.064

Q = 7.296 kJ

Q = 7296 J

Q = m Cp dT

7296 = 108 x 4.18 x (T2 - 29.2)

T2 = 45.36 oC

maximum temperature = 45.4 oC

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