170 mL of 0.25 M HNO3 are mixed with 125 mL of 0.30 M KOH in a calorimeter. If the solutions have an initial temperature of 21.6°C and a molar heat of reaction of –55.78 kJ/mol, then what should the final temperature of the solution be? (the density of the solution is 1.01 g/mL) How can I convert the -55.78 kj/mol to the 2nd temperature? Or I'm doing this incorrectly.I drawn out explanation would be appreciated. thanks (final answer is 23.3*C)

Volume of HNO3 = 170 mL Molarity of HNO3 = 0.25 M Volume of KOH = 125 mL Molarity of KOH = 0.30 M…

Answered: 170 mL of 0.25 M HNO3 are mixed with…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

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170 mL of 0.25 M HNO₃ are mixed with 125 mL of 0.30 M KOH in a calorimeter. If the solutions have an initial temperature of 21.6°C and a molar heat of reaction of –55.78 kJ/mol, then what should the final temperature of the solution be? (the density of the solution is 1.01 g/mL)

How can I convert the -55.78 kj/mol to the 2nd temperature?

Or I'm doing this incorrectly.

I drawn out explanation would be appreciated.

thanks

(final answer is 23.3*C)

Expert Solution

Step 1: Determining number of moles

Volume of HNO₃ = 170 mL

Molarity of HNO₃ = 0.25 M

Volume of KOH = 125 mL

Molarity of KOH = 0.30 M

Initial Temperature (T_i) = 21.6 $° C$

Molar heat of reaction = -55.78 kJ/mole

The chemical equation for neutralization reaction is:

$H N O_{3} + K O H \to K N O_{3} + H_{2} O$

One mole of HNO₃ will neutralize 1 mole of KOH.

Number of moles of HNO₃:

$M o l a r i t y = N u m b e r o f m o l e s \times \frac{1000}{V o l u m e i n m L} N u m b e r o f m o l e s = \frac{M o l a r i t y \times V o l u m e i n m L}{1000} N u m b e r o f m o l e s o f H N O_{3} = \frac{0.25 M \times 170 m L}{1000} N u m b e r o f m o l e s o f H N O_{3} = 0.0425$

Number of moles of KOH:

$M o l a r i t y = N u m b e r o f m o l e s \times \frac{1000}{V o l u m e i n m L} N u m b e r o f m o l e s = \frac{M o l a r i t y \times V o l u m e i n m L}{1000} N u m b e r o f m o l e s o f K O H = \frac{0.30 M \times 125 m L}{1000} N u m b e r o f m o l e s o f K O H = 0.0375$

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