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
I solved the problem and I got 20.44g of chlorine gas
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
i got 6.96g of chlorine gas
am i right?
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.
I don't still get the answer to the question. Please explain more.
You post your work and I'll find the error.
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.
I have seen my mistake. Thanks.
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.
Show working
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%
How give full details of everything
To Samuel--I don't get the remark. Looks detailed to me.
............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.