GCHEM help..

[yakuza]

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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- -->> 5Fe3+ + Mn2+ + 4H2O


Solution: 5 * (25.00 / 1000) * 0.050 * 55.85 * (1 / 1.120) * 100 %


I don't really understand why that 5 comes into play. Also can someone explain the concept behind all this in dat terms?

 

[Danny289]

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- -->> 5Fe3+ + Mn2+ + 4H2O


Solution: 5 * (25.00 / 1000) * 0.050 * 55.85 * (1 / 1.120) * 100 %


I don't really understand why that 5 comes into play. Also can someone explain the concept behind all this in dat terms?

mnO4-======>mn+2 ====> oxid number for "mn" changes from +7 to +2 difference will be 5 in other hand in your equation "5Fe+2" is equal to one mno4- .

 

[keliao]

the setup ofthis solution is kinda confusing....why is it 1/1.120 ???

 

[Contach]

the setup ofthis solution is kinda confusing....why is it 1/1.120 ???

if 55.85 is the molar mass the units are g/mol (im guessing Fe but too lazy to look)
thus to get the amount of moles.. you divide by the amount of grams, in this case 1.120 or you can multiply by the recipricol (aka 1/1.120)

make sense?

 

[chessxwizard]

So first you find out the # of moles of MnO4:
(25/1000) L * 0.05 M = moles MnO4

For every mole MnO4, 5 moles of Fe2+ are consumed:

1 mole MnO4 / (25/1000)*.05 = 5 moles Fe2+ / x
x = moles Fe2+ consumed = 5 * (25/1000) * .05 moles

Convert to grams...

5 * (25/1000) * .05 * 55.84 grams Fe2+

So that's how much Fe2+ was in the ore... but it's asking for the percentage:

[(5 * (25/1000) * .05 * 55.84) / 1.12g sample] * 100 is the answer.

 

[yakuza]

So first you find out the # of moles of MnO4:
(25/1000) L * 0.05 M = moles MnO4

For every mole MnO4, 5 moles of Fe2+ are consumed:

1 mole MnO4 / (25/1000)*.05 = 5 moles Fe2+ / x
x = moles Fe2+ consumed = 5 * (25/1000) * .05 moles

Convert to grams...

5 * (25/1000) * .05 * 55.84 grams Fe2+

So that's how much Fe2+ was in the ore... but it's asking for the percentage:

[(5 * (25/1000) * .05 * 55.84) / 1.12g sample] * 100 is the answer.

Thanks

Will the DAT chem answers be set up like this? Or will they just straight up give it in an easier way

 

[klutzy1987]

This is just achievers stupid way of setting it up, from what i remember the DAT was a little more clear, however there is no gaurantee. They can set it up however they want as long as the numberswork out in the end.

 

[keliao]

if 55.85 is the molar mass the units are g/mol (im guessing Fe but too lazy to look)
thus to get the amount of moles.. you divide by the amount of grams, in this case 1.120 or you can multiply by the recipricol (aka 1/1.120)

make sense?

yeah...thanks