A 74.0 mL sample of 1.0 M NaOH is mixed with 48.0 ml. of 1.0 M H_2SO_4 in a larg

Question

A 74.0 mL sample of 1.0 M NaOH is mixed with 48.0 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 17.2 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 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

Number of moles of NaOH = M*V = 1 M * 0.074 L = 0.074 mol
Number of moles of H2SO4 = M*V = 1 M * 0.048 L = 0.048 mol
1 mol of H2SO4 reacts with 2 mol of NaOH.
So, NaoH is clearly the limiting reagent.
Number of moles of H2SO4 reacted = 0.074/2 =0.037 mol

Heat released per mol H2SO4 = 114000 J
Heat released in this reaction = 0.037*114000 = 4218 J
This heat will raise the temperature of solution.
let final temperature be O oC

Final volume of solution = 74+48 mL = 122 mL
density = 1g/mL
So, mass of solution = density * volume = 122 * 1 =122 g

use:
Q = m*C* (Tf-Ti)
4218 = 122*4.18*(T-17.2)
8.3 = T -17.2
T= 25.5 oC
Answer: 25.5 oC

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