Answer:
71g (assuming experimental data)
Explanation:
The balanced equation for this reaction:
[tex]H_{2} SO_{4}(aq)[/tex] + [tex]2NaOH (s)[/tex] → [tex]Na_{2}SO_{4} (l)[/tex] + [tex]2H_{2} O (l)[/tex]
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 = [tex]\frac{20}{40}[/tex] = 0.5 moles
⇒1 mole of Na2SO4 - 142.04g
∴ 0.5 moles = 142.04 × 0.5
= 71.02g
