Question

# 100.0 mL of 0.800 M aqueous NaOH and 50.00 mL of 0.800 M aqueous H2SO4, each...

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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