Concentration of NaOH(mol/L):1.00
Volume of NaOH used to neutralize acid in sample(ml): 38.88
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2)If the molarity of the acetic acid that was used was 0.830 M calculate your percent error.
1)Use stoichiometry to determine the molarity of the acetic avid(vinegar). NaOH+(HC2H3O2)->H2O+NaC2H3O2 Volume of acid:ml:5.0?
2016-05-31 3:10 am
回答 (2)
2016-05-31 7:17 am
✔ 最佳答案
(1)NaOH + HC₂H₃O₂ → H₂O+Na C₂H₃O₂
Mole ratio NaOH : HC₂H₃O₂ = 1 : 1
No. of moles of NaOH = (1.00 mol/L) × (38.88/1000 L) = 0.03888 mol
No. of moles of HCl = (0.03888 mol) × 1 = 0.03888 mol
Molarity of HC₂H₃O₂ = (0.03888 mol) / (5.0/1000 L) = 7.78 M
(2)
Percent error = [(7.78 - 0.830)/0.830] × 100% = +837%
2016-05-31 7:39 am
NaOH + HC2H3O2 -----------> NaC23O2 + H2O
** The mole of NaOH in 38.88 ml solution : 38.88 ml x (1 M/1000 ml ) = 0.0389 mol
The mole of HC2H3O2 = the mole of NaOH = 0.0389 mol
** The concentration of acid : (0.0389 mol / 5 ml ) x 1000 ml = 7.78 M
The volume of the acid is 5.0 ml ?
** The mole of NaOH in 38.88 ml solution : 38.88 ml x (1 M/1000 ml ) = 0.0389 mol
The mole of HC2H3O2 = the mole of NaOH = 0.0389 mol
** The concentration of acid : (0.0389 mol / 5 ml ) x 1000 ml = 7.78 M
The volume of the acid is 5.0 ml ?
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