Answer to Question #349164 in Chemistry for Grace

Question

140 g of S2Cl2 was collected from the reaction of 640 g of Sulfur and 142 g of Chlorine. How much of the excess reactant was left over?

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:

4 mol of Cl₂ react with 1 mol of S₈

Thus, 2.0 mol of Cl₂ react with:

(2.0 mol Cl₂) × (1 mol S₈ / 4 mol Cl₂) = 0.5 mol S₈

However, initially there is 2.5 mol of S₈ (according to the task)

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

Sulfur (S₈) is excess reactant

Therefore,

(2.5 mol S₈ − 0.5 mol S₈) = 2.0 mol S₈ – excess

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

Therefore,

(2.0 mol S₈) × (256 g S₈ / 1 mol S₈) = 512 g S₈

512 grams of S₈ were left over

Answer: 512 grams of the excess reactant (S₈) was left over