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liq [111]
3 years ago
13

7. Disulfur dichloride can be made by reacting chlorine gas with molten sulfur. What is the yield of S2Cl2 expected in a laborat

ory experiment from the reaction of 10.0 g S8 and 6.00 g Cl2? What if the reaction consistently gives an 85.0 % yield?
Chemistry
1 answer:
Pie3 years ago
3 0

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>
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