Answer to Question #344026 in Chemistry for lyk

Question

140 g of S2Cl2 was collected from the reaction of 640 g of Sulfur and 142 g of Chlorine. What is the maximum amount of S2Cl2 in grams that can be produced from the reaction?

Answer 1

Solution:

The molar mass of sulfur (S₈) is 256 g/mol

The molar mass of chlorine gas (Cl₂) is 71 g/mol

Calculate moles of each reactant:

Moles of S₈ = (640 g S₈) × (1 mol S₈ / 256 g S₈) = 2.5 mol S₈

Moles of Cl₂ = (142 g Cl₂) × (1 mol Cl₂ / 71 g Cl₂) = 2.0 mol Cl₂

Balanced chemical equation:

S₈(l) + 4Cl₂(g) → 4S₂Cl₂(l)

According to stoichiometry:

1 mol of S₈ reacts with 4 mol of Cl₂

Thus, 2.5 mol of S₈ react with:

(2.5 mol S₈) × (4 mol Cl₂ / 1 mol S₈) = 10 mol Cl₂

However, initially there is 2.0 mol of Cl₂ (according to the task)

Therefore, Cl₂ acts as limiting reagent and S₈ is excess reagent

According to stoichiometry:

4 mol of Cl₂ produce 4 mol of S₂Cl₂

Thus, 2.0 mol of Cl₂ produce:

(2.0 mol Cl₂) × (4 mol S₂Cl₂ / 4 mol Cl₂) = 2.0 mol S₂Cl₂

The molar mass of S₂Cl₂ is 135 g/mol

Therefore,

Mass of S₂Cl₂ = (2.0 mol S₂Cl₂) × (135 g S₂Cl₂ / 1 mol S₂Cl₂) = 270 g S₂Cl₂

Mass of S₂Cl₂ = 270 g

Answer: The maximum amount of S₂Cl₂ that can be produced from the reaction is 270 grams