Chemistry

Phosphorus pentachloride is produced by the reaction b/w phosphine and chlorine gas.what mass of chlorine gas would be required to produced 0.015kg of phosphorus pentachloride

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  1. ............2PH3 + 8Cl2 ==> 2PCl5 + 6HCl

    Mols PCl5 needed = grams/molar mass = 15/about 200 = about 0.08. Obviously you need to redo these on your calculator.
    Then convert 0.08 mols PCl5 to mols Cl2 from the equation. That is
    0.08 x (8 mols Cl2/2 mols PCl5) = ? mols Cl2.
    Now convert mols Cl2 to grams Cl2 by grams Cl2 = mols Cl2 x molar mass Cl2.

    Post your work if you get stuck. somewhere.

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  2. How give full details of everything

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    posted by Samuel
  3. To Samuel--I don't get the remark. Looks detailed to me.

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  4. I solved the problem and I got 20.44g of chlorine gas

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    posted by Luke
  5. i got 6.96g of chlorine gas
    am i right?

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    posted by jarvis
  6. I obtained 20.46 g Cl2. There are 2 significant figures in 0.15 kg so 20.46 rounded to 2 places is 20. g.

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  7. I don't still get the answer to the question. Please explain more.

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  8. You post your work and I'll find the error.

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  9. 2PH3 + 8Cl2 ------> 2PCl5 + 6HCl

    0.015kg of PCl5=15g of PCl5


    15g of PCl5*(1 mole/208.24 g)=0.07203 moles of PCl5

    The reaction shows that 2 moles of PCl5=8 moles of Cl2. So,


    0.07203 moles of PCl5*(8 moles of Cl2/2 moles of PCl5)=0.28812 moles of Cl2,


    0.28812 moles of Cl2*(70.906 g/moles)=20 g or 0.020kg of Cl2

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  10. Anonymous is right on.Note that the coefficients are double what they should be (a boo boo I made when I first posted) but that doesn't change anything.

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  11. I have seen my mistake. Thanks.

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  12. If 10g of phosphorus is made to react with 15g of chlorine gas, calculate the percentage yield if 5.5g of phosphorus pentachloride was experimentally produced.

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    posted by Grace
  13. Show working

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    posted by Florence
  14. PH3 ----> PCL5
    33.94g ----> 208.44g
    10g ----> x
    33.94x = 2084.4
    x = 2084.4/33.94
    x = 61.41g of PCL5

    4CL2 ----> PCL5
    284g ----> 208.44g
    15g ----> x
    284x = 3126.6
    x = 3126.6 / 284
    x = 11.01g of PCL5

    Limiting reagent = CL2
    Theoretical Yield = 11.01g
    % yield = actual yield/ theoretical yield * 100
    % yield = 5.5/11.01 * 100
    % yield = 49.95%

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