Using the balanced reaction (HC2H3O2(aq) + KOH(aq)), calculate the amount of 0.0025 M KOH that would neutralize 23 mL of 0.0025 M HC2H3O2??
2018-02-06 1:26 pm
How many moles of salt are produced in the reaction? What is the molar concentration of the salt after the reaction is complete?
回答 (1)
2018-02-06 1:45 pm
Balanced equation for the equation :
HC₂H₃O₂(aq) + KOH(aq) → KC₂H₃O₂(aq) + H₂O(l)
Mole ratio HC₂H₃O₂ : KOH : KC₂H₃O₂ = 1 : 1 : 1
No. of moles of HC₂H₃O₂ reacted = (0.0025 mol/L) × (23/1000 L) = 0.0000575 mol
No. of moles of KOH required = 0.0000575 mol
Volume of KOH needed = (0.0000575 mol) / (0.0025 mol/L) = 0.023 L = 23 mL
No. of moles of the salt KC₂H₃O₂ formed = 0.0000575 mol
Volume of the final solution = (23 + 23) mL = 46 mL = 0.046 L
Molarity of the salt KC₂H₃O₂ formed = (0.0000575 mol) / (0.046 L) = 0.00125 M
HC₂H₃O₂(aq) + KOH(aq) → KC₂H₃O₂(aq) + H₂O(l)
Mole ratio HC₂H₃O₂ : KOH : KC₂H₃O₂ = 1 : 1 : 1
No. of moles of HC₂H₃O₂ reacted = (0.0025 mol/L) × (23/1000 L) = 0.0000575 mol
No. of moles of KOH required = 0.0000575 mol
Volume of KOH needed = (0.0000575 mol) / (0.0025 mol/L) = 0.023 L = 23 mL
No. of moles of the salt KC₂H₃O₂ formed = 0.0000575 mol
Volume of the final solution = (23 + 23) mL = 46 mL = 0.046 L
Molarity of the salt KC₂H₃O₂ formed = (0.0000575 mol) / (0.046 L) = 0.00125 M
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