Question

when magnesium reacts with aluminum chloride, the products are

1. mg + 3cl₂ + 2al
2. mgcl₃ + al
3. mgal + cl₃
4. mg + cl₃ + al
5. mgcl₂ + al
6. mgal + 3cl₂
7. 3mgcl₂ + 2al
8. no reaction

Explanation:

Step1: Identify the reaction type

This is a single - displacement reaction. In a single - displacement reaction, a more reactive metal displaces a less reactive metal from its compound. We need to check the reactivity of Mg and Al. Magnesium (Mg) is more reactive than aluminum (Al) (based on the activity series of metals).

The reactants are magnesium (Mg) and aluminum chloride ($\ce{AlCl_3}$). The general form of a single - displacement reaction is $A + BC
ightarrow AC + B$, where $A$ is the more reactive metal, $B$ is the less reactive metal, and $BC$ is the compound of the less reactive metal.

Step2: Write the balanced chemical equation

The unbalanced equation for the reaction between Mg and $\ce{AlCl_3}$ is:
$\ce{Mg + AlCl_3
ightarrow MgCl_2 + Al}$

To balance the chlorine atoms, we need to find the least common multiple of 3 (from $\ce{AlCl_3}$) and 2 (from $\ce{MgCl_2}$), which is 6. So we put a coefficient of 2 in front of $\ce{AlCl_3}$ and a coefficient of 3 in front of $\ce{MgCl_2}$:
$\ce{Mg + 2AlCl_3
ightarrow 3MgCl_2 + Al}$

Now, to balance the magnesium and aluminum atoms, we put a coefficient of 3 in front of Mg and a coefficient of 2 in front of Al:
$\ce{3Mg + 2AlCl_3 = 3MgCl_2 + 2Al}$

Looking at the options, option 7 is $\ce{3MgCl_2 + 2Al}$, which matches the products of the balanced reaction.

Answer:

1. $\ce{3MgCl_2 + 2Al}$