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.

The balanced chemical reaction is:

S(g) + 3F2(g) ->SF6(g)

We are given the amount of the reactants to be used in the reaction. We use these as the starting values for the calculations. We calculate as follows:

50 g S ( 1 mol / 32.066 g ) = 1.56 mol S
105 g F2 ( 1 mol / 38.00 g ) = 2.76 mol F2

The limiting reactant would be F2 since it will be consumed completely in the reaction. The excess reactant would be S. 

Mass of excess = 1.56 mol - 2.76(1/3) = 0.64 mol S excess

2.00 is the !