Answer to Question #134997 in Chemistry for Dylam

Question

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 atmosphere of chlorine what is the theoritical yield of S2Cl2 in grams, when 4.06 g of S8 is heated with 6.24 g of Cl2? If the actual yield of S2Cl2 is 6.55 g. What is the percent yield?

Answer 1

The balanced reaction equation is the following:

S₈ + 4Cl₂ $\rightarrow$ 4S₂Cl₂.

As one can see from the equation, the number of the moles of sulphur , chlorine and the product disulfide dichloride relate as:

$n(S_8) = \frac{n(Cl_2)}{4} = \frac{n(S_2Cl_2)}{4}$ .

Let's calculate the number of the moles of the reagents:

$n(S_8) = \frac{m(S_8)}{M} = \frac{4.06\text{ g}}{256.52 \text{ g/mol}} = 0.0158 \text{ mol}$ .

$n(Cl_2) = \frac{m(Cl_2)}{M} = \frac{6.24\text{ g}}{70.91 \text{ g/mol}} = 0.0880 \text{ mol}$ .

When one compares the number of the moles of chlorine and sulphur, one can see that 0.0158<0.0880/4. This means that chlorine is in excess. Therefore, it is the number of the moles of the sulphur that must be used to calculate the theoretical yield of the product. The theoretical yield in moles can be calculated using the stoichiometric relation above:

$n(S_2Cl_2) = 4\cdot n(S_8) =0.0158\cdot4 = 0.0633\text{ mol}$.

The molar mass of S₂Cl₂ is 135.04 g/mol. Therefore, its theoretical yield in moles will be:

$m(S_2Cl_2) = n\cdot M = 0.0633\cdot135.04 = 8.55\text{ g}$ .

Finally, the percent yield of the product will be:

$w = \frac{m_{act}}{m_{theor}}\cdot100\% = \frac{6.55\text{ g}}{8.55\text{ g}}\cdot100\% =76.6\%$ .

Answer: When 4.06 g of S₈ is heated with 6.24 g of Cl₂, the theoritical yield of S₂Cl₂ is 8.55 g. If the actual yield of S₂Cl₂ is 6.55 g, the percent yield is 76.6%.