Question

Sulfur hexafluoride is produced by reacting elemental sulfur with fluorine gas. ___ S8(s) + ___ F2(g) ___ SF6(g)
What is the percent yield if 18.3 g SF6 is isolated from the reaction of 10.0 g S8 and 30.0 g F2?

Answer 1

S8 + 24F2 -----> 8SF6

hence its clear that 1 mole of S8 requires 24 moles of F2 and gives 8 mole of SF6.

mole of S8 = 10/256 = 0.039

mole of F2 = 30/38 = 0.789

since 1 mole S8 needs 24 mole of F2

0.039 mole S8 needs 0.039 * 24 = 0.936 mole F2

since mole of F2 = 0.789 <0.936 F2 is the limiting reactant.

now 24 mole of F2 gives 8 moles of SF6

=> 0.789 mole of F2 gives 0.789*8/24 mole of SF6

mole of SF6 = 0.263

hence weight of SF6 = mole * M.wt = 0.263 * 146 = 38.398 g

therefore % of yield = 18.3/38.398 * 100 = 47.658 %