What is the maximum amount in moles of #P_2O_5# that can theoretically be made from 186 g of #P_4# and excess oxygen?

Question

5710 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-28T16:49:42

Well, we need some stoichiometry...

#1/2P_4(s) + 5/2O_2(g) rarr P_2O_5(s)#

#"Moles of phosphorus"# #=# #(186*g)/(31.00*g*mol^-1)=6*mol*"phosphorus atoms"...#...

And so we can make AT MOST #3*mol# #P_2O_5#...

We need #15*mol*"oxygen ATOMS"# for equivalence...

#15*molxx16.00*g*mol^-1=240*g#...

And AT MOST we can make #"THREE MOLES"# of #P_2O_5#...a mass of #3*molxx112*g*mol^-1=336*mol#..

Of course, #P_2O_5#, is actually #P_4O_10#..but we can use this formulation for determination of equivalence...