Question

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . What is the theoretical yield of sodium sulfate formed from the reaction of of sulfuric acid and of sodium hydroxide

Answer 1

Answer:

71g (assuming experimental data)

Explanation:

The balanced equation for this reaction:

$H_{2} SO_{4}(aq)$ + $2NaOH (s)$ → $Na_{2}SO_{4} (l)$ + $2H_{2} O (l)$

Molar mass of H2SO4 = 98.1 g/mol

molar mass of NaOH = 40g/mol

Molar mass of Na2SO4 = 142.04g/mol

⇒ 1 mole or 98.1g of H2SO4 will yield 1 ole of NaSO4; alternately, 2 moles or 49 ×2 = 80g of NaOH produces 1 mole of NaSO4.

Therefore, limiting reactant is NaOH.

Assuming actual experiment is 20g of NaOH,

1 mole - 40g

x moles - 20g = $\frac{20}{40}$ = 0.5 moles

⇒1 mole of Na2SO4 - 142.04g

∴ 0.5 moles = 142.04 × 0.5

= 71.02g