If 12.5 mL of 0.450 mol/L aluminum nitrate reacts with 25.0 mL of 0.380 mol/L sodium hydroxide, what mass of precipitate can be produced?
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Quantitative Chemistry Questiom?
2018-01-22 3:59 pm
回答 (1)
2018-01-22 5:17 pm
Initial number of moles of Al(NO₃)₃ = (0.450 mol/L) × (12.5/1000 L) = 0.00563 mol
Initial number of moles of NaOH = (0.380 mol/L) × (25.0/1000 L) = 0.0095 mol
Al(NO₃)₃(aq) + 3NaOH(aq) → Al(OH)₃(s) + 3NaNO₃(aq)
Mole ratio Al(NO₃)₃ : NaOH(aq) : Al(OH)₃ = 1 : 3 : 1
When 9.5 mol of NaOH completely reacts, Al(NO₃)₃ needed = (0.0095 mol) × (1/3) = 0.00317 mol < 0.00563 mol
Al(NO₃)₃ is in excess, and thus NaOH is the limiting reactant/reagent which completely reacts.
Number of moles of NaOH reacted = 0.0095 mol
Number of moles of Al(OH)₃ produced = (0.0095 mol) × (1/3) = 0.00317 mol
Molar mass of Al(OH)₃ = (27.0 + 16.0×3 + 1.0×3) g/mol = 78.0 g/mol
Mass of Al(OH)₃ produced = (0.00317 mol) × (78.0 g/mol) = 0.247 mol
Initial number of moles of NaOH = (0.380 mol/L) × (25.0/1000 L) = 0.0095 mol
Al(NO₃)₃(aq) + 3NaOH(aq) → Al(OH)₃(s) + 3NaNO₃(aq)
Mole ratio Al(NO₃)₃ : NaOH(aq) : Al(OH)₃ = 1 : 3 : 1
When 9.5 mol of NaOH completely reacts, Al(NO₃)₃ needed = (0.0095 mol) × (1/3) = 0.00317 mol < 0.00563 mol
Al(NO₃)₃ is in excess, and thus NaOH is the limiting reactant/reagent which completely reacts.
Number of moles of NaOH reacted = 0.0095 mol
Number of moles of Al(OH)₃ produced = (0.0095 mol) × (1/3) = 0.00317 mol
Molar mass of Al(OH)₃ = (27.0 + 16.0×3 + 1.0×3) g/mol = 78.0 g/mol
Mass of Al(OH)₃ produced = (0.00317 mol) × (78.0 g/mol) = 0.247 mol
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