What would be the heat released in kJ when 450\ cm^3 of 0.02\ M\ H_2SO_4 is mixed with 650\ cm^3 of 0.1M\ NaOH ? (Heat of neutralization of H_2SO_4 and NaOH is -114\ kJ/mol of H_2SO_4 )

Correct answer: C. 1.026 Number of moles of H_2SO_4 = M × V = 0.02 × 0.450 = 0.009\ mol Number of moles of NaOH = M × V = 0.1 × 0.650 = 0.065\ mol The neutralization reaction is: H_2SO_4 + 2NaOH arrow Na_2SO_4 + 2H_2O From the stoichiometry, 0.009\ mol of H_2SO_4 requires 0.018\ mol of NaOH . Since 0.065\ mol of NaOH is available, H_2SO_4 is the limiting reagent. Heat released per mole of H_2SO_4 is 114\ kJ . Total heat released = 0.009 × 114 = 1.026\ kJ Answer: 1.026

MHT CET 202625 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 25 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Thermodynamics (C)

MHT CET 2026 · 25 April · Morning Shift · Chemistry · Thermodynamics (C)

What would be the heat released in kJ when 450\ cm^3 of 0.02\ M\...

What would be the heat released in kJ when

450\ \text{cm}^3

0.02\ \text{M}\ \text{H}_2\text{SO}_4

is mixed with

650\ \text{cm}^3

0.1\text{M}\ \text{NaOH}

? (Heat of neutralization of

\text{H}_2\text{SO}_4

and

\text{NaOH}

-114\ \text{kJ/mol}

\text{H}_2\text{SO}_4

)

Options

A $0.065$
B $3.71$
C $1.026$
D $1.80$

Correct answer

C. $1.026$

Step-by-step solution

Number of moles of

\text{H}_2\text{SO}_4 = M \times V = 0.02 \times 0.450 = 0.009\ \text{mol}

Number of moles of

\text{NaOH} = M \times V = 0.1 \times 0.650 = 0.065\ \text{mol}

The neutralization reaction is:

\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}

From the stoichiometry,

0.009\ \text{mol}

\text{H}_2\text{SO}_4

requires

0.018\ \text{mol}

\text{NaOH}

. Since

0.065\ \text{mol}

\text{NaOH}

is available,

\text{H}_2\text{SO}_4

is the limiting reagent.

Heat released per mole of

\text{H}_2\text{SO}_4

114\ \text{kJ}

.

Total heat released

= 0.009 \times 114 = 1.026\ \text{kJ}

Answer:

1.026

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