To balance the chemical equation [tex]\( \text{KBrO}_3 + \text{KBr} + \text{H}_2\text{SO}_4 \rightarrow \)[/tex], we need to ensure that there are equal numbers of each type of atom on both sides of the equation. Let's go through the steps to balance this equation.
1. Write the unbalanced equation:
[tex]\[
\text{KBrO}_3 + \text{KBr} + \text{H}_2\text{SO}_4 \rightarrow
\][/tex]
2. Identify the products:
When balancing chemical reactions, it is helpful to predict the products formed. Based on the reactants, the products are likely [tex]\( \text{Br}_2 \)[/tex], [tex]\( \text{K}_2\text{SO}_4 \)[/tex], and [tex]\( \text{H}_2\text{O} \)[/tex].
Therefore, the unbalanced equation is:
[tex]\[
\text{KBrO}_3 + \text{KBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{K}_2\text{SO}_4 + \text{H}_2\text{O}
\][/tex]
3. Balance the bromine (Br) atoms:
We have one bromine atom from [tex]\( \text{KBrO}_3 \)[/tex] and one from [tex]\( \text{KBr} \)[/tex] on the reactant side. Therefore, on the product side, we need to have 2 bromine atoms. However, [tex]\( \text{Br}_2 \)[/tex] contains 2 bromine atoms.
We adjust the coefficients as follows:
[tex]\[
\text{KBrO}_3 + 5\text{KBr} + \text{H}_2\text{SO}_4 \rightarrow 3\text{Br}_2 + \text{K}_2\text{SO}_4 + \text{H}_2\text{O}
\][/tex]
4. Balance the potassium (K) atoms:
On the reactant side, we have one potassium atom from [tex]\( \text{KBrO}_3 \)[/tex] and five potassium atoms from [tex]\( \text{KBr} \)[/tex], giving us a total of six potassium atoms.
On the product side, each [tex]\( \text{K}_2\text{SO}_4 \)[/tex] contains 2 potassium atoms. To balance, we need 3 [tex]\( \text{K}_2\text{SO}_4 \)[/tex].
Now the equation looks like:
[tex]\[
\text{KBrO}_3 + 5\text{KBr} + 3\text{H}_2\text{SO}_4 \rightarrow 3\text{Br}_2 + 3\text{K}_2\text{SO}_4 + \text{H}_2\text{O}
\][/tex]
5. Balance the sulfur (S) atoms:
We have 3 sulfur atoms on the reactant side (from 3 [tex]\( \text{H}_2\text{SO}_4 \)[/tex]).
On the product side, 3 [tex]\( \text{K}_2\text{SO}_4 \)[/tex] also provides 3 sulfur atoms.
6. Balance the oxygen (O) atoms:
For the oxygen atoms, we have:
- 3 oxygen atoms from [tex]\( \text{KBrO}_3 \)[/tex]
- 12 oxygen atoms from 3 [tex]\( \text{H}_2\text{SO}_4 \)[/tex]
Totals to 15 oxygen atoms on the reactant side.
On the product side,
- 12 oxygen atoms come from 3 [tex]\( \text{K}_2\text{SO}_4 \)[/tex]
- 3 oxygen atoms come from 3 [tex]\( \text{H}_2\text{O} \)[/tex]
Totals to 15 oxygen atoms on the product side.
7. Balance the hydrogen (H) atoms:
There are 6 hydrogen atoms on the reactant side (from 3 [tex]\( \text{H}_2\text{SO}_4 \)[/tex]).
On the product side, 3 [tex]\( \text{H}_2\text{O} \)[/tex] provides 6 hydrogen atoms.
With all the elements balanced, the final balanced chemical equation is:
[tex]\[
\text{KBrO}_3 + 5\text{KBr} + 3\text{H}_2\text{SO}_4 \rightarrow 3\text{Br}_2 + 3\text{K}_2\text{SO}_4 + 3\text{H}_2\text{O}
\][/tex]
