Question

what is the product of this combination reaction, keeping in mind that al has a 3+ charge, and cl has a 1 - charge: 2al + 3cl₂→ a 2alcl₃ b 3al + 1cl c 3alcl d 2alcl₂

Explanation:

Step1: Determine the formula of the compound

Al has a \(3+\) charge (\(Al^{3+}\)), Cl has a \(1-\) charge (\(Cl^{-}\)). To form a neutral compound, the ratio of \(Al\) to \(Cl\) should be \(1:3\) (since \(3+( - 1)\times3=0\)), so the formula of the compound is \(AlCl_{3}\).

Step2: Balance the chemical equation

The left - hand side has \(2\) \(Al\) atoms and \(6\) \(Cl\) atoms (\(3Cl_{2}\) gives \(6\) \(Cl\) atoms). For the compound \(AlCl_{3}\), if we take \(2\) moles of \(AlCl_{3}\), it will have \(2\) \(Al\) atoms (\(2\times1\)) and \(6\) \(Cl\) atoms (\(2\times3\)).

Answer:

A. \(2AlCl_{3}\)