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Homework Help: Find the mass of iron used in the experiment.

  1. Jul 2, 2012 #1


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    1. The problem statement, all variables and given/known data
    "A student reacts an unknown mass of iron (Fe) in an excess of sulfuric acid
    (H2SO4). The balanced equation is:
    2 Fe(s) + 3sH2SO4(aq) → Fe2(SO4)3(aq) + 3 H2(g)
    She records the following data:
    Pressure 98.4 kPa
    Volume of dry hydrogen collected 47.3 mL
    Find the mass of iron used in the experiment."

    2. Relevant equations
    PV = nRT and a ratio.

    3. The attempt at a solution
    The solution says (the formatting was ruined when I pasted this and I tried to fix it so if anything is left ruined by accident, sorry):
    " Given: T = 23.0°C + 273 = 296 K; V = 0.0473 L; P = 98.4 kPa;
    R = 8.31 kPa⋅L/mol⋅K, we can calculate the number of moles of hydrogen gas
    produced by the ideal gas equation: n = PV/RT
    n = (98.4 kPa)(0.0473 L)/(8.31 kPa⋅L/mol⋅K)(296 K)
    n = 1.89 x 10^(-3) mol of H2
    From the balanced equation:
    2 Fe(s) + 3H2SO4(aq) → Fe2(SO4)3(aq) + 3 H2(g) hydrogen and iron
    are in a ratio of 3:2 moles. There must, therefore, be 2/3(1.91 x 10^(-3) mol) of Fe
    present, that is, 1.26 x 10^(-3) mol of Fe. Since the molar mass of Fe is 55.85g/mol,
    this represents 0.0705 g of Fe."

    but I don't see how 1.89 x 10^(-3) mol becomes 1.91 x 10^(-3) mol nor do I see why the ratio is 2/3 of approximately the number of moles of H_2; I think it should be 3/2 of exactly the number of moles of H_2 as shown in my work.

    My attempt is attached. I'm feeling that I'm correct but could someone please confirm if I am right and the solution is wrong or tell me why I am wrong if I am wrong?

    Any input would be greatly appreciated!
    Thanks in advance!

    Attached Files:

    Last edited: Jul 2, 2012
  2. jcsd
  3. Jul 2, 2012 #2


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    Science Advisor
    Homework Helper
    Gold Member

    You are asked how many moles of IRON, not hydrogen. You know that the ratio of Fe:H2 is 2:3 and you know the exact value of hydrogen.

    2Fe:3H2 or 2/3 the amount of hydrogen. ie. 2/3Fe:1H2
  4. Jul 3, 2012 #3


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    Isn't that what I did?:

    n_Fe = 3/2 * n_(H_2)

    (as shown in my attached work.)
  5. Jul 3, 2012 #4
    Seems like that its a mistake from where you have copied the solution. If you calculate the 2/3 of 1.89, you get 1.26.
  6. Jul 4, 2012 #5


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    Oh, I got it now. Thanks guys!
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