Identify the limiting reactant when 6.33 g of [tex]H_2SO_4[/tex] reacts with 5.92 g of [tex]NaOH[/tex] to produce [tex]Na_2SO_4[/tex] and water.

H₂SO₄:
m=6.33g M=98g/mol
n = m/M = 6.33g/98g/mol ≈ 0,065mol
NaOH:
m=5.29g M=40g/mol
n = m/M = 5.29g/40g/mol ≈ 0,132mol
2NaOH + H₂SO₄ ⇒ Na₂SO₄ + 2H₂O 2mol : 1mol : 1mol 0.132mol : 0.065mol : 0.065mol too much limiting reactant
Limiting reactant is H₂SO₄.

Answered by Yipes | 2024-06-10

The limiting reactant when 6.33 g of H2SO4 reacts with 5.92 g of NaOH is H2SO4, as we need 0.1290 moles of NaOH for the amount of H2SO4 present, and we have more NaOH than required. ;

Answered by JamesCosmo | 2024-06-24

The limiting reactant when 6.33 g of H₂SO₄ reacts with 5.92 g of NaOH is H₂SO₄, as it is the reactant that will be consumed first based on the stoichiometric ratios. NaOH is in excess. Therefore, the amount of NaOH present is sufficient for the reaction with H₂SO₄.
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Answered by JamesCosmo | 2024-09-27