Question

The amount of NaOH required to neutralize 9.8g H\(_2\)SO\(_4\) is:

• A. 1 mol
• B. 0.1 mol
• C. 2 mol
• D. 0.5 mol

Hint:

For acid-base neutralization reactions, use the stoichiometry to determine the amount of base needed. In this case, 2 moles of NaOH are required to neutralize 1 mole of H\(_2\)SO\(_4\) .

Answer 1

The Correct Option is B

The molar mass of H\(_2\)SO\(_4\) is \(98 \, \text{g/mol}\), and to neutralize H\(_2\)SO\(_4\) , we need 2 moles of NaOH per mole of H\(_2\)SO\(_4\) . The equation for the neutralization reaction is: \[ \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \] Given that we have 9.8g of H\(_2\)SO\(_4\) , the number of moles of H\(_2\)SO\(_4\) is: \[ \text{moles of H\(_2\)SO\(_4\) } = \frac{9.8}{98} = 0.1 \, \text{mol} \] Since 2 moles of NaOH are required to neutralize 1 mole of H\(_2\)SO\(_4\) , the moles of NaOH required are: \[ \text{moles of NaOH} = 0.1 \times 2 = 0.2 \, \text{mol} \] Thus, the correct answer is 0.1 mol of H\(_2\)SO\(_4\) .