The volume of 0.250 M sulfuric acid necessary to react completely with 66.0 g sodium hydroxide is 0.132 L
The reaction of sulphuric acid and sodium hydroxide is expressed as:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
Determine the moles of NaOH present:
Molar mass of NaOH = 23 + 16 + 1 = 40g/mole
Given that mass of NaOH is 66.0g
Mole of NaOH = 66/40 = 1.65moles
Determine the number of moles needed to react with 1.65 moles NaOH:
Moles of H₂SO₄ = 1.65/2 = 0.825 moles
Recall that volume = Molar mass * number of moles
Volume of H₂SO₄ = 0.160 * 0.825
Volume of H₂SO₄ = 0.132L
Hence the volume of 0.250 M sulfuric acid necessary to react completely with 66.0 g sodium hydroxide is 0.132 L
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