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Stuck on 2 Chemistry Questions

  1. Apr 20, 2004 #1
    Hi, I am doing chemistry at my school, and I am currently stuck on a couple of questions that I dont know how to complete. Here are the questions:

    Question 1. An 8.75g. sample of iron wire was dissolved in excess sulfuric acid. All the iron present formed iron (II) ions. This solution was made up to 1.00 L. with distilled water. A 20.0 mL. sample of this solution was pipetted into a conical flask and titrated with 0.025 mol/L potassium permanganate solution. An average titre value of 22.1 mL. of permanganate was required to reach the equivalence point

    a) Calculate (showing all the working) the concentration of iron (II) ions in the 1.00 L. solution if the equation

    for the reaction was
    5Fe2+(aq) + 2MnO4-(aq) + 16H+(aq) = 5Fe3+(aq) + 2Mn2+(aq) + 8H2O(l)

    b) Calculate the % by mass of iron in the original 8.75g sample of wire.

    I am thinking that you need to work out the number of mols of Iron (II) present, using n=m/M, and then use n=CV, and rearrange it into C=n/V and work out the concentration. But I also have a nagging feeling that you need to take into account the mol ratios and use something like n(Fe)/5 = n(MnO4)/2. But I cant use n(Fe), because it is a solid and I need to use m/M for it, and it doesnt have concentration in that equation, which is what I am trying to work out.

    Please, some assistance would be fantastic!

    Question 2. What volume of 0.125 mol/L H2SO4 must be added to 1.25 L of 0.10 mol/L HCl in order to exactly dissolve

    4.5g of Magnesium? (Show all the working out.)
    The equations for the reactions are:

    2HCl + Mg = MgCl2 + H2


    Mg + H2SO4 = MgSO4 + H2


    I tried to work out Question 2 after I wrote this, I think I may have it right. What I did, was I worked out the number of mols of HCl present:

    (I assume you are familiar with n=m/M and n=CV :biggrin: )

    n(hcl)/2 = n(mg)/1 (The division bits are dividing each side by their mole ratio, obtained from the equation)

    Then I began to work out exactly how much Magnesium would actually react with that certain number of mols of HCl:
    (C X V)/2 = m/M (Where m is the mass in grams, M is the molar mass in g/mol)

    Filling in all the known values:
    (0.10 X 1.25)/2 = m/24.31

    After rearranging the equation to get m, I came up with:
    (0.125 X 24.31)/2 = m

    Therefore, m = 1.52g

    So, I found that 1.52g of magnesium will react with the HCl. Now, what is remaining will react with the H2SO4. The

    amount remaining is:

    4.5g-1.52g = 2.98g

    Then, I worked out how much H2SO4 is needed using another stoichiometric calculation:

    n(H2SO4) = n(Mg)

    C X V = m/M

    The remaining Magnesium is used as well to help calculate how much H2SO4 is needed to neutralise what is left of it:
    0.125 X V = 2.98/24.31

    After rearranging:
    V = 0.123/0.125

    V = 0.98L
    I think this question would be all good and right, if not, please correct me. If it is right, can I please have some

    help on Question 1?

    Thanks to anyone who can help!
  2. jcsd
  3. Apr 21, 2004 #2
    An update - it seems i DO have Question 2 right after all - so its just Question 1 that im in the dark about.
  4. Apr 23, 2004 #3

    So anyone doesnt know how to do this?
  5. Apr 23, 2004 #4


    User Avatar
    Science Advisor

    I could give it a shot. I'm not completely sure I understand what is going on in the problem though.

    I'm thinking it could be done backwards. The acid was in excess so just ignore it. For the permanganate, you know the solution concentration and how much it took; find how many moles of permanganate the titration took. Then use the mole ratio to find how much Fe2 was present.

    (x moles MnO4) * (5Fe2/2MnO4) = (y moles Fe2)

    The concentration for the 1L solution should be the same as the 20mL sample. Divide (y moles Fe2) by 0.02L and you should get the proper concentration.

    I don't understand part B.
  6. Apr 24, 2004 #5
    Yeah I actually started to think that this problem isn't so hard after all and it doesn't need too much hard thought at all. But for part B, if you work out the number of mols of Fe2+ used IN THE REACTION, you can divide n by Fe's molar mass to work out the MASS of Fe2+ in the solution that was used up. Then, because that will be a smaller value than the original wire, you can find the percentage by mass by going - m/8.75g * 100, where m is the worked out mass. The reason it is less is because the Fe2+ isnt a pure wire. It has impurities in it.
  7. May 8, 2006 #6

    a) 5Fe2+ + 2MnO4- + 16H+ = 5Fe3+ + 2Mn2+ 8H2O

    No. of moles of MnO4- in 22.1 L= CxV/1000(since the volume is in mL)
    = 22.1 x 0.025/1000
    =5.525 x 10^-4 mol

    Therefore the no. of moles of Fe2+ reacted with MnO4-=5/2 x 5.525 x 10^-4
    (since mole ratio of Fe2+:MnO4-=5:2) =1.38125 x 10^-3mol

    This 1.38125 x 10^-3 mol of Fe2+ is in 20 mL of the solution.
    Therefore, No. of moles of Fe2+= C x V/1000
    1.38125 x 10^-3= C x 20/1000
    C= 0.0690625 mol/L

    b) Fe + H2SO4 = FeSO4 + H2

    In 1L no. of moles of Fe2+ is 0.0690625

    Therefore the no. of moles of Fe is also the same ( from the equation the mole ratio of Fe:Fe2+=1:1)=0.0690625 mol

    No. of moles=Mass/Molar mass
    Mass=No. of moles x Molar mass
    =0.0690625 x 56
    =3.8675 g

    This is the actual mass of Fe present in the 8.75 g of wire.

    Therefore % by mass of Fe present=Actual mass of Fe/Mass of wire x 100
    =3.8675/8.75 x 100

    :blushing: Hope that helped!!!!
    Last edited: May 8, 2006
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