x grams of iron(III) hydroxide (106.9 g/mol) will precipitate if excess iron(III) nitrate is added to 0.850 L of 1.26 M sodium hydroxide?
回答 (1)
2016-06-21 4:52 pm
Fe(NO₃)₃(aq) + 3NaOH(aq) → Fe(OH)₃(aq) + 3NaNO₃(aq)
OR: Mole ratio NaOH : Fe(OH)₃ = 3 : 1
No. of moles of NaOH = (1.26 mol/L) × (0.850 L)
No. of moles of Fe(OH)₃ = (1.26 mol/L) × (0.850 L) × (1/3)
Mass of Fe(OH)₃ = (1.26 mol/L) × (0.850 L) × (1/3) ×(106.9 g/mol) = 38.2 g
OR: Mole ratio NaOH : Fe(OH)₃ = 3 : 1
No. of moles of NaOH = (1.26 mol/L) × (0.850 L)
No. of moles of Fe(OH)₃ = (1.26 mol/L) × (0.850 L) × (1/3)
Mass of Fe(OH)₃ = (1.26 mol/L) × (0.850 L) × (1/3) ×(106.9 g/mol) = 38.2 g
收錄日期: 2021-04-18 15:11:10
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https://hk.answers.yahoo.com/question/index?qid=20160621084553AA6VVal