Asked by Florence

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
Answered by DrBob222
............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.
Answered by Samuel
How give full details of everything
Answered by DrBob222
To Samuel--I don't get the remark. Looks detailed to me.
Answered by Luke
I solved the problem and I got 20.44g of chlorine gas

Answered by jarvis
i got 6.96g of chlorine gas
am i right?
Answered by DrBob222
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.
Answered by Splendid
I don't still get the answer to the question. Please explain more.
Answered by DrBob222
You post your work and I'll find the error.
Answered by Anonymous
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


Answered by DrBob222
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.
Answered by Splendid
I have seen my mistake. Thanks.
Answered by Grace
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.
Answered by Florence
Show working
Answered by Saviour Brown
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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