100.0 mL of 0.800 M aqueous NaOH and 50.00 mL of 0.800 M aqueous
H2SO4, each at 24.0°C, were mixed, see
equation:
2NaOH(aq) + H2SO4(aq) →
Na2SO4(aq) + 2H2O(l)
The final temperature achieved by the solution was 29.3 °C. Neglect
the heat capacity of the cup and the thermometer, and assume that
the solution of products has a density of exactly 1.00 g/mL and a
specific heat capacity of 4.18
J/(g•K). How much heat was absorbed by
the water?
moles of acid H2SO4 = 50 x 0.8 / 1000 = 0.04
moles of NaOH = 100 x 0.8 / 1000 = 0.08
H2SO4 + 2 NaOH ------------------------> Na2SO4 + 2H2O
1 2
0.04 0.08
here both are consumed
moles of water fomred = 0.08
total volume =100 + 50 = 150 mL
mass of solution = volume x density
= 150 x 1
= 150 g
dT = 29.3 -24.0 = 5.3 oC
Q = 150 x 4.18 x 5.3
Q = 3323 J
heat absorbed by the water = 3323 J
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