A 62.0 mL sample of 1.0 M NaOH is mixed with 43.0 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 43.0 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 27.0 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 degree C), and that no heat is lost to the surroundings. The delta H_rxn 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?

Explanation / Answer

The reaction between NaOH and H2SO4 is as follows:

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

Here the total volume = 62.0 ml+ 43.0 ml

= 105 ml

= 0.105 L

Density = mass / volume

Mass = 1.00 g* ml *105 ml

= 105 g

Now calculate the moles of NaOH and H2SO4:

NaOH= Molarity * volume in L

= 1.0* 0.062 l

= 0.062 moles

H2SO4 = 1.0 *0.043

= 0.043 moles

Here NaOHis limiting agent and 0.031 moles of H2SO4 reacted.

Total amount of heat for the neutralization of NaOH with H2SO4 =

-114 KJ/ mole *0.062 mole

= - 7.068 KJ

= -7068 J

the total solution gains the heat given up by the reaction
so
Q = m c (T2-T1)

-7068 J= 105 g *4.18 J / g-C (T2-27.0)

-16.1 = T2-27.0

T2= 10.9 C

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