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# How can I balance this equation?${\text{AlB}}{{\text{r}}_{\text{3}}} + {{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to \_{\text{KBr}} + \_{\text{A}}{{\text{l}}_2}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_3}$

Last updated date: 13th Jun 2024
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Hint: We know that a chemical is called a balanced chemical equation when the number of atoms of all elements is equal in both the product and reactant side of the reaction. While balancing a chemical equation we also need to equate charges if any charge is present.

The given chemical equation is,
${\text{AlB}}{{\text{r}}_{\text{3}}} + {{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to {\text{KBr}} + {\text{A}}{{\text{l}}_2}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_3}$
To balance the K atoms, we have to add coefficient 6 to KBr and 3 to K2SO4. This action equates the number of K atoms in both sides of the reaction.
${\text{AlB}}{{\text{r}}_{\text{3}}} + 3{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to 6{\text{KBr}} + {\text{A}}{{\text{l}}_2}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_3}$
Now, to balance the number of aluminium atoms, we have to add coefficient 2 to ${\text{AlB}}{{\text{r}}_{\text{3}}}$.
${\text{2AlB}}{{\text{r}}_{\text{3}}} + 3{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to 6{\text{KBr}} + {\text{A}}{{\text{l}}_2}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_3}$
Now, in both sides (Product and reactant) of the reaction,
The number of atoms of Al=2
The number of atoms of Br=6
The number of atoms of K=6
The number of atoms of S =3
The number of atoms of O=12
So, the above equation is the balanced chemical equation.

${\text{2AlB}}{{\text{r}}_{\text{3}}} + 3{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to 6{\text{KBr}} + {\text{A}}{{\text{l}}_2}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_3}$