Let's write a step-by-step solution to identify the correctly balanced chemical equation for the reaction between potassium hydroxide (KOH) and sulfuric acid ([tex]\(H_2SO_4\)[/tex]).
1. Identify the Reactants and Products:
The reactants are KOH and [tex]\(H_2SO_4\)[/tex]. These react to produce potassium sulfate ([tex]\(K_2SO_4\)[/tex]) and water ([tex]\(H_2O\)[/tex]).
2. Write the Unbalanced Equation:
[tex]\[
\text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{H}_2\text{O}
\][/tex]
3. Balance the Equation Step-by-Step:
- Balance Potassium (K):
In the unbalanced equation, we have 1 potassium (K) atom on the left side and 2 potassium (K) atoms on the right side. To balance potassium, we need 2 KOH molecules on the reactants side:
[tex]\[
2 \text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{H}_2\text{O}
\][/tex]
- Balance Sulfur (S):
In both reactants and products, we have 1 sulfur (S) atom. So, sulfur is already balanced.
- Balance Oxygen (O):
In the reactants, we have:
- 2 KOH = 2 O atoms
- 1 [tex]\( \text{H}_2\text{SO}_4 \)[/tex] = 4 O atoms
Total = 2 + 4 = 6 O atoms
In the products, we have:
- [tex]\( \text{K}_2\text{SO}_4 \)[/tex] = 4 O atoms
- [tex]\( 2 \text{H}_2\text{O} \)[/tex] = 2 O atoms
Total = 4 + 2 = 6 O atoms
Oxygen atoms are balanced.
- Balance Hydrogen (H):
In the reactants, we have:
- [tex]\( 2 \text{KOH} = 2 \text{H} \)[/tex] atoms
- [tex]\( 1 \text{H}_2\text{SO}_4 = 2 \text{H} \)[/tex] atoms
Total = 2 + 2 = 4 H atoms
In the products, we have:
- [tex]\( 2 \text{H}_2\text{O} = 4 H \)[/tex] atoms
Hydrogen atoms are balanced.
Given that each side of the equation now has an equal number of atoms for each element, we have the balanced chemical equation:
[tex]\[
2 \text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2 \text{H}_2\text{O}
\][/tex]
So, the correctly balanced chemical equation for the reaction of KOH and [tex]\(H_2SO_4\)[/tex] is:
[tex]\[
2 \text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2 \text{H}_2\text{O}
\][/tex]
