Question

A mixture of 1.0 moles of Al and 3.0 mole of Cl₂ are allowed to react as:

$2Al(s) + 3Cl_2(g) \rightarrow 2AlCl_3(g)$

Then moles of excess reagent left unreacted is:

• A. 3.5
• B. 1.0
• C. 1.5
• D. 2.5

Answer 1

Answer: (C)

1.5

Answer 2

Step 1: Determine molar ratio.
Reaction requires 2 mol Al per 3 mol Cl₂ → for every 1 mol Al, need $\tfrac{3}{2}=1.5$ mol Cl₂.

Step 2: Check Cl₂ needed vs available.
Available Cl₂ = 3.0 mol
Cl₂ needed to consume 1.0 mol Al = 1.0 × 1.5 = 1.5 mol

Step 3: Calculate excess.
Excess Cl₂ = Available Cl₂ − Needed Cl₂ = 3.0 − 1.5 = 1.5 mol

Conclusion: Al is the limiting reagent; Cl₂ left unreacted = 1.5 mol.