Question

what is the formula equation for the reaction between sulfuric acid and dissolved sodium hydroxide if all products and reactants are in the aqueous or liquid phase?
hcl(aq) + naoh(aq) → nacl(s) + h2o(l)
h2so4(aq) + ca(oh)2(aq) → caso4(aq) + 2h2o(l)
h2so4(aq) + 2naoh(aq) → na2so4(aq) + 2h2o(l)
2h3po4(aq) + 3ca(oh)2(aq) → ca3(po4)(s) + 6h2o(l)

Explanation:

Step1: Identify reactants

Sulfuric acid is $\mathrm{H_2SO_4}$ and sodium - hydroxide is $\mathrm{NaOH}$.

Step2: Determine reaction type

It's an acid - base neutralization reaction. Acid + base → salt + water.

Step3: Write products

$\mathrm{H_2SO_4}$ reacts with $\mathrm{NaOH}$ to form sodium sulfate ($\mathrm{Na_2SO_4}$) and water ($\mathrm{H_2O}$).

Step4: Balance equation

For sulfuric acid and sodium hydroxide reaction, 1 mole of $\mathrm{H_2SO_4}$ reacts with 2 moles of $\mathrm{NaOH}$ to give 1 mole of $\mathrm{Na_2SO_4}$ and 2 moles of $\mathrm{H_2O}$, so the balanced equation is $\mathrm{H_2SO_4}(aq)+ 2\mathrm{NaOH}(aq)
ightarrow\mathrm{Na_2SO_4}(aq)+ 2\mathrm{H_2O}(l)$.

Answer:

C. $\mathrm{H_2SO_4}(aq)+ 2\mathrm{NaOH}(aq)
ightarrow\mathrm{Na_2SO_4}(aq)+ 2\mathrm{H_2O}(l)$