Q. \( \text{H}_2\text{SO}_4 + \text{NaOH} \)

Q: Is \mathrm{H_2SO_4} diprotic, and how does that affect stoichiometry with \mathrm{NaOH} ?

\mathrm{H_2SO_4} is diprotic. It needs 2 moles of \mathrm{NaOH} per 1 mole of \mathrm{H_2SO_4} to fully neutralize.

Q: What products form when \mathrm{H_2SO_4} reacts with excess \mathrm{NaOH} ?

Complete neutralization produces \mathrm{Na_2SO_4} and \mathrm{H_2O} . Any extra \mathrm{NaOH} remains unreacted.

Q: How do you determine which reactant is limiting?

Convert given masses to moles, then use the ratio \mathrm{H_2SO_4} : \mathrm{NaOH} = 1:2 . The reactant requiring more of the stoichiometric partner is excess; the other is limiting.

Answer

To balance the reaction between sulfuric acid and sodium hydroxide, first write the unbalanced equation:

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

Count atoms: H requires 2 waters’ worth, so use 2 \(\text{NaOH}\). This gives \(2\) Na and the sulfate balances automatically.

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

Final balanced equation: \( \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)

Detailed Explanation

We interpret the problem as a neutralization reaction between sulfuric acid and sodium hydroxide.

Step 1: Write the reactants.

The reactants are:

\[ \text{H}_2\text{SO}_4 + \text{NaOH} \]

Step 2: Identify the acid-base reaction pattern.

Sulfuric acid is diprotic, meaning it can donate two protons. Sodium hydroxide provides hydroxide ions that accept those protons.

Step 3: Write the products you expect.

For a neutralization between an acid and a base, the typical products are:

\[ \text{salt} + \text{water} \]

The sulfate ion pairs with sodium ions to form sodium sulfate, and the protons combine with hydroxide to form water.

Step 4: Balance the reaction.

Start by balancing each element.

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

Now check atoms:

Hydrogen on the left comes from \(\text{H}_2\text{SO}_4\) and from \(\text{NaOH}\). There are 2 H already from \(\text{H}_2\text{SO}_4\), so we need 2 more H from two \(\text{NaOH}\) molecules to make two water molecules.
Sulfur and oxygen must match accordingly.

To make \(\text{Na}_2\text{SO}_4\), we need 2 sodium ions, so we use 2 molecules of \(\text{NaOH}\).

Write the balanced equation:

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

Step 5: Verify the balanced equation.

Check each element:

Sulfur (S): \(1\) on both sides.
Hydrogen (H): Left has \(2\) from \(\text{H}_2\text{SO}_4\) plus \(2\) from \(2\text{NaOH}\), totaling \(4\). Right has \(2\) from \(2\text{H}_2\text{O}\), totaling \(4\).
Oxygen (O): Left has \(4\) from \(\text{H}_2\text{SO}_4\) plus \(2\) from \(2\text{NaOH}\), totaling \(6\). Right has \(4\) from \(\text{Na}_2\text{SO}_4\) plus \(2\) from \(2\text{H}_2\text{O}\), totaling \(6\).
Sodium (Na): Left has \(2\) from \(2\text{NaOH}\). Right has \(2\) in \(\text{Na}_2\text{SO}_4\).

Step 6: Final answer (balanced chemical equation).

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

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General Chemistry FAQs

What is the balanced equation for the reaction of \( \mathrm{H_2SO_4} \) with \( \mathrm{NaOH} \)?

\[ \mathrm{H_2SO_4 + 2\,NaOH \rightarrow Na_2SO_4 + 2\,H_2O} \]

Is \( \mathrm{H_2SO_4} \) diprotic, and how does that affect stoichiometry with \( \mathrm{NaOH} \)?

\( \mathrm{H_2SO_4} \) is diprotic. It needs \(2\) moles of \( \mathrm{NaOH} \) per \(1\) mole of \( \mathrm{H_2SO_4} \) to fully neutralize.

What products form when \( \mathrm{H_2SO_4} \) reacts with excess \( \mathrm{NaOH} \)?

Complete neutralization produces \( \mathrm{Na_2SO_4} \) and \( \mathrm{H_2O} \). Any extra \( \mathrm{NaOH} \) remains unreacted.

What happens if \( \mathrm{NaOH} \) is limiting (not enough base)?

Only partial neutralization occurs. The reaction may stop at \( \mathrm{NaHSO_4} \) if \(1\) equivalent of \( \mathrm{NaOH} \) is used per mole of \( \mathrm{H_2SO_4} \).

How do you determine which reactant is limiting?

Convert given masses to moles, then use the ratio \( \mathrm{H_2SO_4} : \mathrm{NaOH} = 1:2 \). The reactant requiring more of the stoichiometric partner is excess; the other is limiting.

What is the net ionic equation for the neutralization?

\[ \mathrm{H^+ + OH^- \rightarrow H_2O} \] With sulfuric acid, it happens twice overall: \( \mathrm{2H^+} \) reacts with \( \mathrm{2OH^-} \).

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