Q122 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Chemistry
What would be the heat released in kJ when $450\ \text{cm}^3$ of $0.02\ \text{M}\ \text{H}_2\text{SO}_4$ is mixed with $650\ \text{cm}^3$ of $0.1\text{M}\ \text{NaOH}$? (Heat of neutralization of $\text{H}_2\text{SO}_4$ and $\text{NaOH}$ is $-114\ \text{kJ/mol}$ of $\text{H}_2\text{SO}_4$)
- A$0.065$
- B$3.71$
- Correct answerC$1.026$
- D$1.80$
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, $0.009\ \text{mol}$ of $\text{H}_2\text{SO}_4$ requires $0.018\ \text{mol}$ of $\text{NaOH}$. Since $0.065\ \text{mol}$ of $\text{NaOH}$ is available, $\text{H}_2\text{SO}_4$ is the limiting reagent.
Heat released per mole of $\text{H}_2\text{SO}_4$ is $114\ \text{kJ}$.
Total heat released $= 0.009 \times 114 = 1.026\ \text{kJ}$
Answer: $1.026$
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, $0.009\ \text{mol}$ of $\text{H}_2\text{SO}_4$ requires $0.018\ \text{mol}$ of $\text{NaOH}$. Since $0.065\ \text{mol}$ of $\text{NaOH}$ is available, $\text{H}_2\text{SO}_4$ is the limiting reagent.
Heat released per mole of $\text{H}_2\text{SO}_4$ is $114\ \text{kJ}$.
Total heat released $= 0.009 \times 114 = 1.026\ \text{kJ}$
Answer: $1.026$
Hence, the correct answer is option: 'C'.