How many mL of 0.4M sulfuric acid solution should be used to neutralize 8 gr of NaOH?
回答 (2)
2018-01-11 5:47 pm
Molar mass NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol
No. of moles of NaOH reacted = (8 g) / (40 g/mol) = 0.2 mol
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Mole ratio H₂SO₄ : NaOH = 1 : 2
No. of moles of H₂SO₄ used = (0.2 mol) × (1/2) = 0.1 mol
Volume of H₂SO₄ used = (0.1 mol) / (0.4 mol/L) = 0.25 L = 250 mL
No. of moles of NaOH reacted = (8 g) / (40 g/mol) = 0.2 mol
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Mole ratio H₂SO₄ : NaOH = 1 : 2
No. of moles of H₂SO₄ used = (0.2 mol) × (1/2) = 0.1 mol
Volume of H₂SO₄ used = (0.1 mol) / (0.4 mol/L) = 0.25 L = 250 mL
2018-01-11 7:54 pm
One.
收錄日期: 2021-04-24 00:53:30
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180111092927AABiRRO