Question

One beaker contains 20 mL of 1.60 M HCI, and a second beaker contains 60 mL of 0.40 M NaOH. Both solutions were initially at 20°C. We pour both beakers into a large insulated container and monitor the temperature. The molar enthalpy of neutralization is-57.3 kJ/mol. What is the highest temperature reached? 5.
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Answer #1

moles of HCl = 20 x 1.60 / 1000 = 0.032 mol

moles of NaOH = 60 x 0.40 / 1000 = 0.024 mol

mass of solution = 40 + 20 = 60 g

HCl   +    NaOH    ----------------> NaCl   +   H2O

1               1                                    1             1

0.032       0.024                          

here limiting reagent is NaOH.

Q = n x delta H

= 0.024 x 57.3

Q = 1.375 kJ

Q = m Cp dT

1375 = 80 x 4.184 x (Tf - 20 )

Tf - 20 = 4.108

Tf = 24.1

final temperature = 24.1 oC

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