Answer:
11.4g of S₂Cl₂ is the expected yield
9.69g of S₂Cl₂ are produced with a 85% yield
Explanation:
The reaction of sulfur S₈ with Cl₂ to produce S₂Cl₂ is:
S₈ + 4Cl₂ → 4S₂Cl₂
<em>Where 1 mole of sulfur reacts with four moles of chlorine to produce four moles of disulfur dichloride.</em>
To find the limiting reactant you need to convert mass of each reactant to moles using its molar mass, thus:
S₈ (Molar mass: 256.52g/mol): 10.0g ₓ (1mol / 256.52g) = 0.0390 moles S₈
Cl₂ (Molar mass: 70.9g/mol): 6.00g ₓ (1mol / 70.9g) = 0.0846 moles Cl₂
For a complete reaction of 0.0390 moles of sulfur, there are necessaries:
0.0390 mol S₈ ₓ (4 mol Cl₂ / 1 mol S₈) = <em>0.156 moles Cl₂. </em>As you have just 0.0846 moles of chlorine, Cl₂ is the limiting reactant.
As 4 moles of Cl₂ produce 4 moles of S₂Cl₂.<em> 0.0846 moles of Cl₂ produce, in theory, 0.0846 moles of S₂Cl₂ (Molar mass: 135.04g/mol). </em>In mass:
0.0846 moles S₂Cl₂ ₓ (135.04g/mol) =
<h3>11.4g of S₂Cl₂ is the expected yield</h3>
If you produce just the 85.0% of yield, mass of S₂Cl₂ is:
11.4g ₓ 85% =
<h3>9.69g of S₂Cl₂</h3>
