Question

chem help

Answer 1

@matt101

Answer 2

ig et .02845L MnO4-*(0.2025mol/L)*(5mol Fe^2+/mol MnO4-)*(56g/mol)=.1613g of Fe^2+
%Fe=(.1613g/2.893g)=5.58%
is this correct?

Answer 3

What's the concentration of the permanganate solution?

Answer 4

the c0ncentrati0n might be in this questi0n if 26.23m of potassium permanganate solution is required to titrate 1.041 g f ferrus amm0nium sufate hexahydrate, FeSO4(NH4)2SO46H2O, CALCULATE the molarity of the KMnO4 solution.

for that i got MnO4-+4H2O +Fe2+-->MnO2+2H2O+4OH-+5Fe3+
1mol KMnO4 oXIDIZES 5mol Fe2+
1.041g=2.65*10^3mol Fe2+
5.3*10^-4mol/0.02623L=0.02M

Then i went on to answer the question i needed help on.

Answer 5

Is it 26.23 M or 26.23 m? The capitalization makes a difference

Answer 6

Ok that makes sense - one more question though: are you sure that's the right chemical reaction? You have water on both sides

Answer 7

i think so. my book is'nt to clear but it looks like how they want me to do it.

Answer 8

Alright I'll work with what we have. I got the same concentration as you did for the first question. For the second part, I got 0.15932 g of iron (I think you used the wrong concentration in your calculation), which works out to 5.51%.

But yes you did it right!

Answer 9

great. i'll have to recheck then to make sure

Answer 10

@matt101 can you walk me through the titration pic. i need to fill in the blanks and i'm looking through my book and it has no formulas on how to do it.

Answer 11

Well the moles of iron you can calculate from the mass of FAS (moles of iron = moles of FAS). Moles of KMNO4 is a fifth of that according to the chemical reaction. Molarity you can then calculate by dividing that by the volume of KMNO4 used. For the last row just average the concentrations and then find the standard deviation of those values.

Similar process for the bottom half - basically you're following the same process you went through to answer your second question above, just with your measured values instead.

Answer 12

0kay so for my calculations of:
moles of iron present for sample #1 i got 1.117, sample#2 ig ot 1.117 and for sample#3 i got 1.117
Moles of KMnO4 present i got for sample#1 .0200,sample#2 .02,sample#3.0200
Molarity of KMnO4 solution sample#1 .6897M, SAMPLE#2 .625M,sample#3 .6987M
mean molarity and average deviation i got .0118

For the unkown:
Moles of KMnO4 present sample#1 .0089661,sample#2 .00875,sample#3 .0174
moles of iron present sample#1 .0448305,sample#2 .04375,sample#3.087
Mass of iron sample#1 2.504,sample#2 2.443,sample#3 .9718
% of iron present sample#! 786.4,sample#2 768.94,sample#3 314.3
Mean % iron present and average deviation 3.745%

This is what i have when i did trhe calculations to fill in the blanks. Not sure if they are right @matt101 could you double check these for me?

Answer 13

Hello @ash90 !

You probably didn't notice, but you posted your \(\large\sf\color{blue}{\underline{\href{ /study#/groups/chemistry}{Chemistry}}}\) question in the Biology section.
Next time when you ask a question, check if you're in the right section :)

Answer 14

I'll go through the first sample. For the first part:

Moles of iron = moles of FAS = 1.0215 g / 392 g/mol = 2.6 x 10^(-3) mol
Moles of KMNO4 = moles of iron/5 = 2.6 x 10^(-3) mol / 5 = 5.2 x 10^(-4) mol
Molarity of KMNO4 solution = 5.2 x 10^(-4) mol / 0.029 L = 0.018 M

Repeat this for the other two samples and use the values to find the mean and standard deviation.

For the second part:

Volume of KMNO4 used was 0.013 L
Moles of KMNO4 = 0.013 L x 0.018 M = 2.34 x 10^(-4) mol
Moles of iron = moles of KMNO4 x 5 = 2.34 x 10^(-4) mol x 5 = 1.17 x 10^(-3) mol
Mass of iron = 1.17 x 10^(-3) x 56 g/mol = 0.066 g
Percent of iron present = 0.066 g / 0.3184 g = 0.207 = 20.7%

Repeat this for the other two samples and use the values to find the mean and standard deviation.

Answer 15

okay after doing it the way you showed i get for my calculations:
moles of iron present for sample #1 i got 2.6*10^-3mol, sample#2 i g ot 2.9*10^-3mol and for sample#3 i got 2.6*10^-3mol
Moles of KMnO4 present i got for sample#1 5.2*10^-4mol,sample#2 5.8*10^-4mol,sample#3 5.2*10^-4mol
Molarity of KMnO4 solution sample#1 0.018M, SAMPLE#2 0.018M,sample#3 0.018M
mean molarity and average deviation i got 0.049

For the unkown:
Moles of KMnO4 present sample#1 2.34*10^-4mol,sample#2 2.52*10^-4mol,sample#3 4.54*10^-4mol
moles of iron present sample#1 1.17*10^-3mol,sample#2 1.26*10^-3mol,sample#3 2.27*10^-3mol
Mass of iron sample#1 0.066g,sample#2 0.071g,sample#3 0.127g
% of iron present sample#! 20.7%,sample#2 22.2%,sample#3 41.1%
Mean % iron present and average deviation 0.0127%

@matt101

Answer 16

Looks good, just your means and deviations are two separate things so you should have 2 separate numbers.

Answer 17

okay and thanks so much for being patient with me. and taking the time to help me.

Answer 18

Not a problem - glad I was able to help!