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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?

Verified Answer :

roytucker

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%

The theoretical yield of disulfide dichloride is 8.53 grams.

The percent yield of the reaction is 76.8%.

Explanation:

Given:

The reaction between 4.06 grams of octasulfur and 6.24 grams of chlorine gas give 6.24 grams of disulfide dichloride.

To find:

The theoretical yield and percent yield of the reaction.

Solution:

The mass of octasulfur= 4.06 g

The molar mass of octasulfur=\(\frac{4.06 g}{256.52 g/mol}=0.0158 mol\)

The mass of chlorine gas = 6.24 g

The moles of chlorine gas =\(\frac{6.24 g}{70.906 g/mol}=0.0880 mol\)

\(S_8(l) + 4Cl_2 (g) \rightarrow 4S_2Cl_2(l)\)

According to reaction,1 mole of octasulfur reacts with 4 moles of chlorine gas, then 0.0158 moles of octasulfur will react with:

\(=\frac{4}{1}\times 0.0158 mol=0.0632 \text{mol of}Cl_2\)

But we are having 0.0880 moles of chlorine gas and only 0.0632 moles of chlorine gas will be used in a reaction which means that the chlorine gas is present in an excess amount, hence an excessive reagent and octasulfur is limiting reagent.

The theoretical yield of disulfide dichloride will depend upon the moles of octasulfur.

According to the reaction, 1 mole of octasulfur gives 4 moles of disulfide  dichloride then 0.0158 moles octasulfur will give:

\(=\frac{4}{1}\times 0.0158 mol=0.0632 \text{mol of}S_2Cl_2\)

The mass of 0.0632 moles of disulfide dichloride:

\(=0.0632 mol\times 135.04 g/mol=8.53 g\)

The theoretical yield of disulfide dichloride = 8.53 g

The theoretical yield of disulfide dichloride is 8.53 g

The actual yield of disulfide dichloride = 6.55 g

The percent yield of the reaction:

\(Yield(\%)=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100\\Yield=\frac{6.55 g}{8.53 g}\times 100=76.8\%\)

The percent yield of the reaction is 76.8%.

Learn more about theoretical yield and percent yield.

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