Question

100.0 mL of 0.800 M aqueous NaOH and 50.00 mL of 0.800 M aqueous
H_{2}SO_{4}, each at 24.0°C, were mixed, see
equation:

2NaOH(aq) + H_{2}SO_{4}(aq) →
Na_{2}SO_{4}(aq) + 2H_{2}O(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?

Answer #1

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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