How many grams of barium sulfate are produced if 25.00mL of 0.225 M BaCl2 completely react given the reaction:?
回答 (2)
2016-12-13 12:29 pm
Molar mass of BaSO₄ = (137.3 + 32.1 + 16×4) g/mol = 233.4 g/mol
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)
OR: Mole ratio BaCl₂ : BaSO₄ = 1 : 1
No. of moles of BaCl₂ reacted = (0.225 mol/L) × (25/1000 L) = 0.005625 mol
No. of moles of BaSO₄ produced = 0.005625 mol
Mass of BaSO₄ produced = (0.005625 mol) × (233.4 g/mol) = 1.31 g
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)
OR: Mole ratio BaCl₂ : BaSO₄ = 1 : 1
No. of moles of BaCl₂ reacted = (0.225 mol/L) × (25/1000 L) = 0.005625 mol
No. of moles of BaSO₄ produced = 0.005625 mol
Mass of BaSO₄ produced = (0.005625 mol) × (233.4 g/mol) = 1.31 g
2016-12-13 12:31 pm
(0.02500 L) x (0.225 mol/L BaCl2) x (1 mol BaSO4 / 1 mol BaCl2) x (233.3896 g BaSO4/mol) =
1.31 g BaSO4
1.31 g BaSO4
收錄日期: 2021-04-20 17:51:18
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