Gc Question From Achiever

[vDDmaniaC]

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Hi, guys.
I was going over DAT achiever that I took today, and there is a problem that I could't still solve. I read the solution over and over. I am not still sure which value is which.

THe problem is

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 %

Can anyone goes over step by step so I can fully absorb this problem?

THanks a bunch

 

[Contach]

Hi, guys.
I was going over DAT achiever that I took today, and there is a problem that I could't still solve. I read the solution over and over. I am not still sure which value is which.

THe problem is

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 %

Can anyone goes over step by step so I can fully absorb this problem?

THanks a bunch

The reasoning here is simple, but its a lot of numbers so take it step by step. It's much like a titration, so it will involve stoichiometry.
First start with number of mols of Mn04 = n = cv = (0.050M)(25.00mL)/1000mL/L
now times this by the stoic ratio of Mn04: 5Fe2+ = 5*(0.050M)(25.00mL)/1000mL/L
now you have the mols of Fe2+ that will be needed. but you have to find the amount of grams of Fe2+ in this ore (ore means its some mixture of metals that will have some pure Fe2+ in it).
So you take the mols of Fe2+ * molar mass = 5*(0.050M)(25.00mL)/1000mL/L*55.85g/mol and this will give you grams of Fe2+
Now you need to find the percent by mass of Fe2+ in the ore. so its mass-of-Fe2+/total-mass x 100%
5*(0.050M)(25.00mL)/1000mL/L*55.85g/mol*(1/1.20)*100%

This is choice D.

Hope this is right!

 

[vDDmaniaC]

Thanks, that really helped me a lot!

 

[yakuza]

oh my freaking goodness...

I know how to do the problem and can get the exact answer but I do it a completely different way(not the traintrack dimensional analysis thing)
and can't match my answer+work to the answer choices..

Is the DAT going to have answer choices set up this way?