How do you write the equation for this reaction: Aluminum bromide and chlorine gas react to form aluminum chloride and bromine gas?

Md Ariful Islam

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In , the must add together to become $0$.

In aluminum bromide, aluminum has an oxidation number of $+3$ and bromine has an oxidation number of $-1$. In order to sum to zero, you need to have THREE bromines, with an overall charge/oxidation number of $-3$, and one aluminum, with a charge/oxidation number of $+3$, since $-3+3=0$.

This is why one reactant is $"AlBr"_3$.

The other reactant, $"Cl"_2$, needs two chlorines, as it is one of the that requires two of itself to have an oxidation state/charge of zero.

This is a single replacement reaction, in which the free element in the reactants becomes part of the compound in the products, replacing one of the original elements in the compound, and makes the replaced element become the free element in the products.

Bromine is the replaced element and becomes $"Br"_2$, the free element in the products. It is $"Br"_2$ since it is another element that needs two of itself to have a charge/oxidation number of zero.

The oxidation number of chlorine is $-1$, and for the same reason as mentioned in paragraph one, three chlorines are needed to produce an oxidation number/charge of zero with aluminum, resulting in the compound $"AlCl"_3$.

The second equation given is balanced by mass, but if you don't need a balanced equation, use the first equation.

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