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[CHEM] Simple stoichiometry problem help

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data

    5.00g alloy of Magnesium and Aluminum is treated with excess HCl, forming MgCl2 and AlCl3 and 6.65L of H2 at 25 degrees celcius and 99.2kPa. What is the mass percent of Mg in the alloy?

    2. Relevant equations

    PV = nRT

    3. The attempt at a solution

    Using PV = nRT, I found the moles of H2 gas to be approximately 0.2663. Now I attempted to setup my overall reaction by combining the half reactions.

    Mg + 2HCL --> MgCl2 + H2 (1)
    2Al + 6HCL --> 2AlCl3 + 3H2 (2)

    Adding them we obtain:

    Mg + 2Al + 8HCL ---> MgCl2 + 2AlCl3 + 4H2

    Using the above equation, I know that 1 mol of Mg reacts with 4 moles of H2, therefore x moles of Mg reacts with 0.2663 moles of H2. Therefore, the moles of Mg is approximately equal to .2663/4 = 0.066575 mol.

    Multiplying the moles of Mg by it's molar mass, 24.31 g/mol gives you about 1.6g of Mg, which in turn if 32% of the 5g compound.

    The problem is that the answer to the question isn't 32%, it's 16%. If it's 16% then my mole ratio was likely wrong because if you have a ratio of 1 mol of Mg for every 8 moles of H2, you get 16% as your answer, I just don't see how that ratio would come to be.

    Furthermore, the charges are not balanced in my equations, is that a problem?

    - Thanks
  2. jcsd
  3. Oct 15, 2007 #2


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    STOP THERE. You can't do this. You can't add them together since you don't know the ratio of Mg to Al. As you have shown it, it is 1:2 but is this the only ratio possible?
  4. Oct 15, 2007 #3
    Hmm, then I've no idea how to work with those two separate equations to obtain an answer.
  5. Oct 15, 2007 #4


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    Your equation is good, take a look at the left side of it, you will see that 1 atom Mg take
    2 molecules HCl with a release of 2 H forming quickly H2 so 2/8 of the genesis from H2 is the responsibilty from the Mg. In the calculation you get 0.066575 mol H2 so than divide this amount by 2 multiply by atom weight Mg and you get your amount of Mg further calculation gives 16%. In brief: one mole Mg release two mole H atoms with the formation of one mole H2
  6. Oct 19, 2007 #5


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    There is a lot more that you know that you haven't thought about. For example,

    Let x = grams Al and let y = grams Mg, then

    Since y=5-x, we know the mass of both Mg and Al in terms of one unknown, 'x'.

    Another thing that you know,

    moles Mg = (5-x) grams Mg / FW Mg and,
    moles Al = x grams Al / FW Al.

    That should help you figure it out...
  7. Oct 19, 2007 #6


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    This is not true always.
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