Question

A solution contains \(9.8 \, \text{g}\) of \(H_2SO_4\). How much NaOH is required to completely neutralize it? (molar mass of NaOH = \(40 \, \text{g mol}^{-1}\))

• A. \(0.4 \, \text{g} \)
• B. \(0.2 \, \text{g} \)
• C. \(8 \, \text{g} \)
• D. \(1.2 \, \text{g} \)
• E. \(1.6 \, \text{g} \)

Hint:

Always balance the chemical equation first, then apply mole ratio. For diprotic acids like \(H_2SO_4\), 1 mole reacts with 2 moles of base.

Answer 1

The Correct Option is C

Concept: Neutralization reaction between acid and base: \[ H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O \] Key idea:
• Use mole concept: \( \text{moles} = \frac{\text{given mass}}{\text{molar mass}} \)
• Stoichiometry from balanced equation

Step 1: Calculate moles of \(H_2SO_4\).
Molar mass of \(H_2SO_4 = 98 \, \text{g/mol}\) \[ \text{Moles of } H_2SO_4 = \frac{9.8}{98} = 0.1 \, \text{mol} \]

Step 2: Use stoichiometric ratio.
From equation: \[ 1 \, \text{mol } H_2SO_4 \Rightarrow 2 \, \text{mol } NaOH \] \[ 0.1 \, \text{mol } H_2SO_4 \Rightarrow 0.2 \, \text{mol } NaOH \]

Step 3: Convert moles of NaOH to mass.
\[ \text{Mass} = 0.2 \times 40 = 8 \, \text{g} \]