Question

Sulfur and fluorine react in a combination reaction to produce sulfur hexafluoride: S(g) + 3F2(g) ->SF6(g)

If 50 g S is allowed to react as completely as possible with 105.0g F2(g), what mass of the excess reactant is left.

Answer 1

What I did so far:
• I determine which one is the limiting reagent by calculating the theoretical yield of SF6.

50g S(1 mol S/32.07gS)(1mol SF6/1mol S)(146.07g SF6/1 mol SF6) = 227.84 g SF6

105 g F2(1 mol F2/38 gF2)(1mol SF6/3mol F2)(146.07g SF6/1 mol SF6) = 134.38 g SF6

So Sulfur is the excess reactant... then ? what is the next step?

Answer 2

Yes Sulfur would be the excess reactant. Your next step would be determining how much of your limiting reactant (F2) reacts with your excess reactant (S) by doing this then you would be able to determine how much of the sulfur is used and then you subtract that from the original. To determine this you would set up your equation much like you have already done to determine which one is in excess.

Answer 3

okay. I'll try give me one sec. :)

Answer 4

I got 29.54 g S, then subtract that by 50 g...it will be 20.46 or 20.5 g S!

Thank you so much!

Answer 5

Yep and that is what I got when I went through it, no problem!