[narutodemonkill]

I  tried this question multiple times , and get the right
answer but still think my method is wrong.


The illustration to the left represents a mixture of
phosphorus (orange) and fluorine (green) molecules.

If the molecules in the above illustration react to form PF3
according to the equation
P4 + 6F2 -> 4PF3 I counted 7 fluorine and 1 phosphorus.

So to figure out the limiting reagent I used the ratio
method  (*** would dividing molecules by coefficient also give you limiting reagent? not sure if this only works with mols.)

7 Fluorine molecules *( 4 molecules PF3/6 molecules of
fluroine= 4.67 molecules of PF3 produced.

than I did 1 molecule of P4 *( 4 molecules of PF3/ 1
molecule of P4)= 4 molecules of PF3.

From this I concluded that P4 was the limiting reagent.

to figure out how many molecules of F2 where in excess I
worked backwards.

4 molecules of PF3 *( 6 molecules of F2/4 molecules of
PF3)=6 molecules of F2 used up and therefore 1 molecule is
in excess.

is there a different way to solve this. And is this even a
correct method.


**If they asked for mols of F2 would it give you same answer? not sure if mols and molecules are interchangeable in this case?
because co-efficent refers to molecules and mols.

so answer could also be 1 mol of F2 is in excess?

[Borek]

Hard to say if you are 100% right not seeing the picture, but it looks like you are right - P₄ is a limiting reagent, there is one molecule of F₂ in excess.

Mols and molecules are to some extent interchangeable - please read http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations