FeCl3(aq) + 3 NaOH(aq) → Fe(OH)3(s) + 3 NaCl(aq)
When 10.00 mL of a NaOH solution reacts with excess iron(III) chloride, 2.48 grams of iron(III) hydroxide precipitates. What is the molarity of the sodium hydroxide solution?
A beaker contains 125.0 mL of a 0.813 M NaOH solution. 12.00 mL of 0.0871 M HCl is added to the beaker and the solution is mixed. Determine the molarity of hydroxide ion, [OH‾], in this final solution.
I know that M=N/V but I don't understand how to solve with all the extra information that's given.
Finding Molarity (Chemistry)?
2020-09-08 4:56 pm
回答 (1)
2020-09-08 6:06 pm
1.
Molar mass of Fe(OH)₃ = (55.8 + 16.0×3 + 1.0×3) g/mol = 106.8 g/mol
FeCl₃(aq) + 3 NaOH(aq) → Fe(OH)₃(s) + 3 NaCl(aq)
Mole ratio NaOH : Fe(OH)₃ = 3 : 1
Moles of Fe(OH)₃ = (2.48 g) / (106.8 g/mol) = 0.02322 mol
Moles of NaOH = (0.02322 mol) × 3 = 0.06966 mol
Molarity of NaOH = (0.06966 mol) / (10.00/1000 L) = 6.97 M
OR:
(2.48 g Fe(OH)₃) × (1 mol Fe(OH)₃ / 106.8 g Fe(OH)₃) × (3 mol NaOH / 1 mol Fe(OH)₃) / (10.00/1000 L NaOH solution)
= 6.97 M NaOH
====
2.
Initial moles of NaOH = (0.813 mol/L) × (125.0/1000 L) = 0.101625 mol
Initial moles of HCl = (0.0807 mol/L) × (12.00/1000 L) = 0.000968 mol
Balanced equation for the reaction:
NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(ℓ)
Mole ratio NaOH : HCl = 1 : 1
Since (Initial moles of NaOH) > (Initial moles of HCl),
NaOH is in excess, and HCl is the limiting reactant.
Moles of HCl reacted = 0.000968 mol
Moles of NaOH reacted = 0.000968 mol
Moles of NaOH left unreacted = (0.101625 - 0.000968) mol = 0.100657 mol
Volume of the final solution = (125.0 + 12.00 mL) = 137 mL = 0.137 L
Molarity of NaOH in the final solution = (0.100657 mol) / (0.137 L) = 0.735 M
Each mole of NaOH dissociate to give 1 mole of OH⁻ ion.
Molarity of OH⁻ ion in the final solution = 0.735 M
Molar mass of Fe(OH)₃ = (55.8 + 16.0×3 + 1.0×3) g/mol = 106.8 g/mol
FeCl₃(aq) + 3 NaOH(aq) → Fe(OH)₃(s) + 3 NaCl(aq)
Mole ratio NaOH : Fe(OH)₃ = 3 : 1
Moles of Fe(OH)₃ = (2.48 g) / (106.8 g/mol) = 0.02322 mol
Moles of NaOH = (0.02322 mol) × 3 = 0.06966 mol
Molarity of NaOH = (0.06966 mol) / (10.00/1000 L) = 6.97 M
OR:
(2.48 g Fe(OH)₃) × (1 mol Fe(OH)₃ / 106.8 g Fe(OH)₃) × (3 mol NaOH / 1 mol Fe(OH)₃) / (10.00/1000 L NaOH solution)
= 6.97 M NaOH
====
2.
Initial moles of NaOH = (0.813 mol/L) × (125.0/1000 L) = 0.101625 mol
Initial moles of HCl = (0.0807 mol/L) × (12.00/1000 L) = 0.000968 mol
Balanced equation for the reaction:
NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(ℓ)
Mole ratio NaOH : HCl = 1 : 1
Since (Initial moles of NaOH) > (Initial moles of HCl),
NaOH is in excess, and HCl is the limiting reactant.
Moles of HCl reacted = 0.000968 mol
Moles of NaOH reacted = 0.000968 mol
Moles of NaOH left unreacted = (0.101625 - 0.000968) mol = 0.100657 mol
Volume of the final solution = (125.0 + 12.00 mL) = 137 mL = 0.137 L
Molarity of NaOH in the final solution = (0.100657 mol) / (0.137 L) = 0.735 M
Each mole of NaOH dissociate to give 1 mole of OH⁻ ion.
Molarity of OH⁻ ion in the final solution = 0.735 M
收錄日期: 2021-05-01 09:50:58
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