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Mg metal reacts with HCl to produce hydrogen gas?

  1. Aug 7, 2016 #1
    1. The problem statement, all variables and given/known

    Mg metal reacts with HCl to produce hydrogen gas. What is the minimum volume of HCl solution (27% by weight) required to produce 12.1 g of H2? Density of HCl solution is 1.14 g/cm3


    2. Relevant equations

    This question is related to stoichiometry, Molarity of solution




    3. The attempt at a solution


    I am not getting what exactly this question asks about. This is very confusing can anyone help me solving this
     
  2. jcsd
  3. Aug 7, 2016 #2

    Ygggdrasil

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    Always start with a balanced chemical equation.
     
  4. Aug 7, 2016 #3
    This is balanced chemical equation Mg + 2HCl -------> H2 + MgCl2
    but what to do next
     
  5. Aug 7, 2016 #4

    Ygggdrasil

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    Can you figure out the number of moles of HCl required to produce 12.1g of H2?
     
  6. Aug 8, 2016 #5
    Yes As per balanced chemical equation if 1 mole of hydrogen require 2 mole of HCl then 6.05 Moles of Hydrogen will require 6.05×2 Moles of HCl i.e 12.1 Moles.By multiplying it with 22.4dm3 doesn't give the right answer
    The answer to this question should come to be 1423cm^3
     
  7. Aug 8, 2016 #6

    epenguin

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    Don't just say hydrogen as you won't be sure what you mean. It's 6.05 moles of H2.

    Actually for this calculator you could just forget about that and just say it's 12.1 moles of H .

    So 12.1 moles of HCl.. How many grams of HCl is that? (Your 22.4 dm3 comes from something different, calculation is about gases which are not involved in this question).
     
    Last edited: Aug 8, 2016
  8. Aug 8, 2016 #7
    All right so how to do this? Do you have any idea?
     
  9. Aug 8, 2016 #8
    21.1×36.5=770.15g
     
  10. Aug 8, 2016 #9

    Ygggdrasil

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    22.4 dm3/mole applies only to gasses, as epenguin mentioned above.

    The rest of the problem is just dimensional analysis, figuring out how many moles per liter are in a 27% (w/w) solution of HCl. 27% (w/w) means that for every 100g of solution, you have 27g of HCl. From that information (and the density), you should be able to figure out how many moles of HCl you have per liter of solution.
     
  11. Aug 8, 2016 #10

    epenguin

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    Have you mistaken 12.1 for 21.1?

    Then for the volume you could think of the calculation as not chemistry, but arithmetic of type that you were probably doing around age 10.
     
  12. Aug 9, 2016 #11
    My mistake . I got the answer. So 6.05 moles of H2 require 12.1 moles of HCl
    12.1mol×36.5g/mol= 441.65g Now dividing it by w/w percentage as given 441.65÷0.27=1635.74 Density = m/V or V= Density×m Therefore V=1864.7cm^3 This is the answer I got but I have a question that why do we divide the mass by w/w percentage?
    :biggrin:
     
    Last edited: Aug 9, 2016
  13. Aug 9, 2016 #12

    epenguin

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    The chemistry is finished, it only remains to find what volume of the given solution contains 441.65 g but you are having difficulty with arithmetic of masses and volumes, usually learnt around age 10. (Not uncommon you may be relieved to hear.)

    You need to render yourself confident and independent of help for this kind of arithmetical calculation which is often asked and a frequent necessity in practice. If you can't do the required calculation one trick* is try the same calculation for a simplified version of the problem. E.g. suppose you had to get 500g of HCl from a solution that was 0.5g/g by weight and the density of the mixture was say 2 g/ml how would you do that?

    Secondly you need to practice the habit of habitualLy checking your calculations for qualitative and rough quantitative reasonableness. E.g you have about 1600 g of the HCl-water mixture . But this is denser than 1g/ml so the HCl is contained in less than 1600 ml instead of more as you have.

    By avoidable mistakes like this you can easily do most of the work right and get only a little credit. (At my school we were told if you noticed your exam answer unreasonable and couldn't find and correct the mistake in time, to just make a note that the result was unreasonable.)

    And although book answers are not infallible and we have often enough here found mistakes in them, the big difference between your answer and the book one you quoted earlier is a warning to be careful.

    Using your figures I calculated a volume of 1434.86 ml. This is close to the book answer, but not identical. Probably the discrepancy arises through using the same approximate atomic masses you did and it looks like the difference between these and the accurate ones is of about the right amount. So when you realise how to calculate this, use the accurate 1.00794 for H and 35.453 for Cl.

    *This is useful for far more advanced work than the present, and is found is a short, cheap, popular and very helpful book "How to solve it" by Polya.
     
    Last edited: Aug 9, 2016
  14. Aug 9, 2016 #13

    Ygggdrasil

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    27% (w/w) means that you have 0.27 g of HCl per 1g of solution. Therefore, the conversion is: $$441.65 \textrm{ g HCl} * \frac{1\textrm{ g solution}}{0.27\textrm{ g HCl}} = 1635.74\textrm{ g solution}$$
     
  15. Aug 9, 2016 #14
    Thank you for help and sharing tips of solving problems also pointing out mistakes. Thank you once again.:oldsmile:
     
    Last edited: Aug 10, 2016
  16. Aug 10, 2016 #15
    Now I got it . Thank you for clearing my doubts:oldsmile:
     
  17. Aug 10, 2016 #16

    epenguin

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    Did you get the same answer as the book? Could you set out this last part of the calculation, so we can mark this probem as solved?
     
  18. Aug 10, 2016 #17
    Mg + 2 HCl → MgCl2 + H2

    (12.1 g H2) / (2.01588 g H2
    /mol) x (2 mol HCl / 1 mol H2) x (36.4611 g HCl/mol) = 437.7 g HCl

    (437.7 g HCl) / (0.27) = 1621.11 g HCl solution

    (1621.11 g) / (1.14 g/cm^3) = 1422 cm^3
    In my book answer is 1423.
     
  19. Aug 11, 2016 #18

    epenguin

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    OK similar things will occur as part of other problems as you go on, so hope you are well set for them now.
     
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