A student determined the heat of neutralization of sulfuric acid mixed with sodi
ID: 941259 • Letter: A
Question
A student determined the heat of neutralization of sulfuric acid mixed with sodium hydroxide solution. 35.0 mL of 1.00 M H2SO4 were added to 70.5 mL of 1.00 M NaOH solution.
a) Calculate the mass of the reaction mixture. Assume the density of the mixture is 1.03 b/mL.
b) Calculate the number of moles of H2SO4 and NaOH that react when the solutions are mixed.
c) If change in temperature is 7.83 degrees C, calculate the heat transferred from the reaction. Assume that the specific heat of the mixture is 3.89 J/gdeg.
Explanation / Answer
a)
total volume = 35 + 70.5 = 105.5 mL
density = mass / volume
mass = density x volume
= 1.03 x 105.5
= 108.665 grams
b)
no of moles = Moolarity x volume in liters
no of moles of H2SO4 = 1 x 0.035
= 0.035 moles
no of moles of NaOH = 1 x 0.0705
= 0.0705 moles
c)
H = m Cp dT
= 108.665 grams x 3.89 J/gC x 7.83 ºC
= 3309.7946 J = 3.309 kJ
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