Question

PARTILDAJA DATA Heinz Mass of empty flask Mass of flask + vinegar Molarity of NaOH solution - 0 Initial buret reading Final buret reading (Red tope) Flask #1 89.5699 g 3.9289 100 OM 1.01 m2 38.04 mL Flask N2 89.3769 92.8414 0.100OM 1.32 ml 38.54mL Flask #3 99.375g 103:8 589 0.1000M 1.62 ml 39.94 ml CHEMICAL REACTION Write a balanced chemical equation for the titration reaction: HC H202 + Nook Hoo NacQH, Oz tog DATA ANALYSIS 4 4.369 0 27.03 m2 _9.701 mol .465 a 4.4859 27.29 mm 37.62 mL 2.68 mol 9.65 8 noul Mass of vinegar mole Volume of NaOH used Moles of NaOH used Moles HC2H302 neutralized Mass of HC2H302 60.06g 60.06 60.06g % HC2H302 in vinegar Average % HC2H302 in vinegar Accepted value (filled in by the instructor) 184

Answer 1

1) moles of NaOH = molarity×volume/1000

For flask 1 : moles of NaOH = 0.1×37.03/1000 =0.003703

For flask 2 : moles of NaOH = 0.1×37.22/1000 =0.003722

For flask 3 : moles of NaOH = 0.1×37.62/1000 = 0.003762

2) moles of HC2H3O2 = moles of NaOH

For flask 1 : moles of HC2H3O2 = 0.003703

For flask 2 : moles of HC2H3O2 = 0.003722

For flask 3 : moles of HC2H3O2 = 0.003762

3) Mass of HC2H3O2 = moles × molar mass

For flask 1 : mass = 0.003703×60.06 = 0.222 g

For flask 2 : mass = 0.003722×60.06 = 0.224 g

For flask 3 : mass = 0.003762×60.06 = 0.226 g

4) % HC2H3O2 in vinegar = mass of HC2H3O2×100/vinegar mass

For flask 1 : % HC2H3O2 = 0.222×100/4.369 = 5.08%

For flask 2 : % HC2H3O2 = 0.224×100/4.465 = 5.02%

For flask 3 : % HC2H3O2 = 0.226×100/4.483 = 5.04%

5) Average % = (5.08+5.02+5.04)/3 = 15.14/3 = 5.05%

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