Anonymous Chegg expert answered this Was this answer helpful? 1 0 8,605 answers
ID: 503716 • Letter: A
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Anonymous Chegg expert answered this
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8,605 answers
The complete reaction:
2MnO4- (aq) + 16H+ (aq) + 5C2O42- (aq) --> 2Mn2+ (aq) + 8H2O (l) + 10CO2 (g)?
then:
for
[KMnO4] = 0.0195 M --> [MnO4-] = 0.0195 M
so:
Mass of Green Salt = experimentally measured, i.e. not calculated
Initial V in buret = experimentally measured, i.e. not calculated
Final V = experimentally measured, i.e. not calculated
Net V of KMnO4 = Vfinal - Vinitial
From your datta; all is correct
Moles of C2O4-2 calculation:
You will need to first:
calculate moles of KMnO4 = MnO4-
mol of MnO4- = M*V = 0.0915*V
choose Trial 1:
mol of MnO4- = M*V = 0.0195*(11.849) = 0.2310555 mmol of MnO4-
relate to C2O4-2
2:5 ratio
0.2310555-->5/2*0.2310555= 0.57763875 mmol of C2O4-2 = 0.00057763875
which is correct
then
Incorrect:
mass of salt in trial = 0.0933 g
moles of C2O4-2 = 0.00057763875 mol
so:
mass of Salt per mol of C2O4 = 0.0933/5/0.00057763875 = 32.3 g per mol
This is incorrect, I can't get where you got 162,91 g / mol
sa (Gen Chem) Name KESULTS and CALCULATIONS DUE Record data and calculations in your lab notebook. The following tables may be used as a guide for h to organize your data and calculations. of several trials are carried out, only the full calculations for one trial are required to be shown.) Analysis of oxalate in the Complex Salt -Rough Run" Trial 1 Trial 2 Trial 3 Trial 4 Blank Mass of Green Salt og 33 o 93 og 37 og 34. used Final Buret Volume Net KMnos volume Moles of C20 g Green Salt/mol croi Average g Green Salt mol czo.3 Suspected outlier ISS. S67 Final Average g Green Salt/mol Co? Io Sul. 12.00 LExplanation / Answer
Volume of KMnO4 used = 11.849 mL
moles of KMnO4 = 0.0195 M* 0.011849L = 2.31*10^-4 moles
2 moles KMNO4 reacts with 5 moles Oxalate. So, number of moles of oxalate present in green salt
= (5/2) * 2.31*10^-4 moles
= 5.776*10^-4 moles
moles of C2O4^2- per gram = 0.0933gm/ 5.776*10^-4 moles = 161.52 moles/gm
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