Disulfide dichloride (S2Cl2) is used in the vulcanization of rubber, a process that prevents the slippage of rubber molecules past one another when stretched. It is prepared by heating sulfur in an atmosphere of chlorine:

S8 (l) + 4Cl2 (g)  4S2Cl2 (l)

What is the theoretical yield of S2Cl2 in grams when 4.06 g of S8 are heated with 6.24 g of Cl2? If the actual yield of S2Cl2 is 6.55 g, what is the percent yield?

First step Convert each of the masses into moles. I'm going to assume you know how to do that so you have 0.0158 mol of S8 and 0.088 mol of Cl2.

Now look at the equation. The mole ration is 1 mole of S8 to 4 mole of Cl2. Now compare the mole you just calculated. 0.088 mol of Cl2 is NOT 4 times as much as 0.0158 mol of S8. Therefore there is not enough mol of S8 to completely use up the Cl2. That means that the reaction will proceed until all the S8 is used up and then stop. That makes S8 the limiting reagent.

Now, for every mole of S8 used you get 4 mol of S2Cl2. Since 0.0158 mol of S8 will be completel used up that means that (4 x 0.0158) or 0.0632 mol of S2Cl2 will be formed. Convert that to mass by multiplying 0.0632 mol x 135.02 g/mol of S2Cl2. that gives 8.53 g of theoretical yield.

Since you are only getting 6.55 g of S2Cl2 then the percent yield is 6.55/8.53 = 0.767 or 76.7%