Question

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. \(3.71\)
• C. \(1.026\)
• D. \(1.80\)

Hint:

Find the moles of acid and base, decide which is the limiting reagent, and multiply by the heat of neutralization per mole of acid.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The heat released depends on the limiting reagent. One mole of \(\text{H}_{2}\text{SO}_{4}\) needs two moles of \(\text{NaOH}\).

Step 2: Find the moles
\(\text{H}_{2}\text{SO}_{4}\): \(0.450\ \text{L}\times 0.02\ \text{M} = 0.009\ \text{mol}\).
\(\text{NaOH}\): \(0.650\ \text{L}\times 0.1\ \text{M} = 0.065\ \text{mol}\).

Step 3: Limiting reagent
0.009 mol of acid needs only 0.018 mol of base, so the acid is limiting.

Step 4: Calculation
Heat released \(= 0.009\times 114 = 1.026\ \text{kJ}\).

Final Answer:
The heat released is 1.026 kJ. \[ \boxed{\text{(C) }1.026} \]