Disulfide dichloride \(\left(\mathrm{S}_{2} \mathrm{Cl}_{2}\right)\) 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:$$\mathrm{S}_{8}(l)+4 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{~S}_{2} \mathrm{Cl}_{2}(l)$$What is the theoretical yield of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) in grams when \(4.06 \mathrm{~g}\) of \(\mathrm{S}_{8}\) are heated with \(6.24 \mathrm{~g}\) of \(\mathrm{Cl}_{2} ?\) If the actual yield of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) is \(6.55 \mathrm{~g}\), what is the percent yield?

Short Answer

Expert verified
The theoretical yield of \(S_2 Cl_2\) is 8.53 g, and the percent yield is 76.78%.

Step by step solution

01

Understand the Balanced Chemical Equation

The given balanced chemical equation is \(\mathrm{S}_{8}(l)+4 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{S}_{2} \mathrm{Cl}_{2}(l)\). It indicates that 1 mole of \(\mathrm{S}_{8}\) reacts with 4 moles of \(\mathrm{Cl}_{2}\) to produce 4 moles of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\).
02

Calculate the Moles for Each Reactant

We are given the mass of \(\mathrm{S}_{8}\) and \(\mathrm{Cl}_{2}\) as \(4.06 \mathrm{~g}\) and \(6.24 \mathrm{~g}\) respectively. We use molar mass to convert these to moles, which gives us moles of \(\mathrm{S}_{8} = 4.06 \mathrm{~g} / 256.52 \mathrm{~g/mol} = 0.0158 \mathrm{~mol}\) and moles of \(\mathrm{Cl}_{2} = 6.24 \mathrm{~g} / 70.90 \mathrm{~g/mol} = 0.0880 \mathrm{~mol}\)
03

Identify the Limiting Reactant

We calculate the needed amount of \(\mathrm{Cl}_{2}\) for the given amount of \(\mathrm{S}_{8}\) to react completely, using the stoichiometry from the balanced equation: \((0.0158 \mathrm{~mol} \mathrm{S}_8) * (4 \mathrm{~mol} \mathrm{Cl}_2 / 1 \mathrm{~mol} \mathrm{S}_8) = 0.0632 \mathrm{~mol} \mathrm{Cl}_2\). Since the amount of provided \(\mathrm{Cl}_2\) (0.0880 mol) is more than needed, \(\mathrm{S}_8\) is the limiting reactant.
04

Calculate Theoretical Yield

We use the balanced chemical equation again to calculate the theoretical yield of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) in grams from the limiting reactant \(\mathrm{S}_{8}\): \((0.0158 \mathrm{~mol} \mathrm{S}_8) * (4 \mathrm{~mol} \mathrm{S}_2 \mathrm{Cl}_2 / 1 \mathrm{~mol} \mathrm{S}_8) * 135.04 \mathrm{g/mol} = 8.53 \mathrm{g}\) of \(\mathrm{S}_2 \mathrm{Cl}_2\)
05

Calculate Percent Yield

We use the formula for percent yield, which is \( \text{actual yield} / \text{theoretical yield} * 100%\). It gives us \(6.55 \mathrm{g} / 8.53 \mathrm{g} * 100% = 76.78%\)

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