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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 L

Explanation / 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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