chemistry question??

[Denro]

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What percentage of iron is present in the ore, if a 1.120 g sample requires 25.00 ml of 0.050 M KMnO4 to fully oxidize all of the dissolved Fe2+?
8H+ + 5Fe2+ + MnO4- à 5Fe3+ + Mn2+ + 4H2O




A. 5 * (1000 / 25.00) * 0.050 * 55.85 * (1 / 1.120) * 100 %
B. 5 * (1000 / 25.00) * 0.050 * 55.85 * 1.120 * 100 %
C. 5 * (0.050 / 1000) * 25.00 * 55.85 * 1.120 * 100 %
D. 5 * (25.00 / 1000) * 0.050 * 55.85 * (1 / 1.120) * 100 %
E. 5 * (25.00 / 1000) * 0.050 * 55.85 * 1.120 * 100 %

this seems like it should be easy.. but im going blank on this.. can somebody help me with this problem.. thanks.

 

[joonkimdds]

8H+ + 5Fe2+ + MnO4- à 5Fe3+ + Mn2+ + 4H2O

looks kinda wierd to me.

 

[chordata]

What percentage of iron is present in the ore, if a 1.120 g sample requires 25.00 ml of 0.050 M KMnO4 to fully oxidize all of the dissolved Fe2+?
8H+ + 5Fe2+ + MnO4- à 5Fe3+ + Mn2+ + 4H2O




A. 5 * (1000 / 25.00) * 0.050 * 55.85 * (1 / 1.120) * 100 %
B. 5 * (1000 / 25.00) * 0.050 * 55.85 * 1.120 * 100 %
C. 5 * (0.050 / 1000) * 25.00 * 55.85 * 1.120 * 100 %
D. 5 * (25.00 / 1000) * 0.050 * 55.85 * (1 / 1.120) * 100 %
E. 5 * (25.00 / 1000) * 0.050 * 55.85 * 1.120 * 100 %

this seems like it should be easy.. but im going blank on this.. can somebody help me with this problem.. thanks.

I'll guess D

 

[Denro]

ur right.. but how do u get to it??

 

[chordata]

ur right.. but how do u get to it??

What percentage of iron is present in the ore, if a 1.120 g sample requires 25.00 ml of 0.050 M KMnO4 to fully oxidize all of the dissolved Fe2+?
8H+ + 5Fe2+ + MnO4- à 5Fe3+ + Mn2+ + 4H2O




D. 5 * (25.00 / 1000) * 0.050 * 55.85 * (1 / 1.120) * 100 %

25ml/1000ml = 0.025L
0.05mol/L * 0.025L = 0.00125 mols KMnO4
0.00125 mol * 5 = 0.00625 mols Fe
0.00625 mol Fe * 55.85 GMW Fe = 0.349 g Fe oxidized
0.349/1.12 * 100% = percent iron present

 

[joonkimdds]

What percentage of iron is present in the ore, if a 1.120 g sample requires 25.00 ml of 0.050 M KMnO4 to fully oxidize all of the dissolved Fe2+?
8H+ + 5Fe2+ + MnO4- à 5Fe3+ + Mn2+ + 4H2O




D. 5 * (25.00 / 1000) * 0.050 * 55.85 * (1 / 1.120) * 100 %

25ml/1000ml = 0.025L
0.05mol/L * 0.025L = 0.00125 mols KMnO4
0.00125 mol * 5 = 0.00625 mols Fe
0.00625 mol Fe * 55.85 GMW Fe = 0.349 g Fe oxidized
0.349/1.12 * 100% = percent iron present

where is KMnO4 ?

 

[Denro]

What percentage of iron is present in the ore, if a 1.120 g sample requires 25.00 ml of 0.050 M KMnO4 to fully oxidize all of the dissolved Fe2+?
8H+ + 5Fe2+ + MnO4- à 5Fe3+ + Mn2+ + 4H2O




D. 5 * (25.00 / 1000) * 0.050 * 55.85 * (1 / 1.120) * 100 %

25ml/1000ml = 0.025L
0.05mol/L * 0.025L = 0.00125 mols KMnO4
0.00125 mol * 5 = 0.00625 mols Fe
0.00625 mol Fe * 55.85 GMW Fe = 0.349 g Fe oxidized
0.349/1.12 * 100% = percent iron present

I can see your train of thought.. but i thought GEW was molar mass/ N where n is the number of hydrogens and OH.. so i cant see how 1.12 grams fits the GEW eq???

 

[chordata]

I can see your train of thought.. but i thought GEW was molar mass/ N where n is the number of hydrogens and OH.. so i cant see how 1.12 grams fits the GEW eq???

I'm not sure I understand your question...
The 1.12 grams is used to determine the percent composition.
When you convert the moles KMnO4 used to Fe moles consumed you then convert the moles of Fe to its equivalent weight by multiplying it by its GMW. This weight represents the amount of Fe in the orignal ore compound. You take this weight and divide it by the original 1.12 g and then multiply it by 100% to get the percent composition.
Hope this helps...

 

[dlink]

Don't confuse yourself, Denro. You've been given molarity (M) of KMnO4 and NOT its normality (N). But since you brought it out here, take a look of the less frequently used approach, delienated as follows:

Normality of KMnO4, N = aM
where a = change in oxidation # of MnO4- to Mn2+ = |2-7|=5
N = 5 x 0.05 equiv/liter

Now, # of equivalents of KMnO4, e = VN
where V = volume of KMnO4 = 25.00 ml = (25.00/1000) liter
e = (25.00/1000) x 5 x 0.05 equiv ------ (1)

The fundamental concept of equivalents and normal solutions implies that # of equivalents of KMnO4 used equals the # of equivalents of Fe in the ore sample [which is e as arrived in (1)] . In this reaction, the oxidation # of Fe increases only by one unit, leaving the equivalent weight being the same as its atomic weight, 55.85:

? g Fe = (1) x 55.85 = 5 x (25.00/1000) x 0.05 x 55.85 ----- (2)

Percentage of iron present in the ore is therefore:

[(2)/1.120] x 100% = 5 x (25.00/1000) x 0.05 x 55.85 x (1/1.120) x 100%

Will this help clarify or is it confusing you even more? Interesting chem question posted, isn't it? 🙂