A 62.0 mL sample of 1.0 M NaOH is mixed with 55 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 55 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 20.5 degree C. 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 *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/mol H_2SO_4. What is the maximum measured temperature in the Styrofoam cup? Number 3.74 degree C

Explanation / Answer

moles of NaOH = 62 x 1 / 1000 = 0.062

moles of H2SO4 = 55 x 1 / 1000 = 0.055

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

2                1

0.062        0.055

here limiting reagent is NaOH. so we consider the moles of NaOH to calculate heat

delta H = - Q / n

- 114 = - Q / 0.062 x 10^3

Q = 7068 J

mass of solution = 62 + 55 = 117 g

cp = 4.184

Q = m Cp dT

7068 = 117 x 4.18 x (Tf - 20.5)

Tf = 34.95 oC

maximum temperature = 34.95 oC

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