To determine which chemical equation is balanced, we need to ensure the number of atoms for each element is the same on both sides of the equation. Let's analyze each option step-by-step.
Option A:
[tex]\[ \text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{KHSO}_4 + \text{H}_2\text{O} \][/tex]
- On the left side:
- K: 1
- O: 5 (1 from KOH + 4 from [tex]\( \text{H}_2\text{SO}_4 \)[/tex])
- H: 3 (1 from KOH + 2 from [tex]\( \text{H}_2\text{SO}_4 \)[/tex])
- S: 1
- On the right side:
- K: 1
- O: 5 (4 from [tex]\( \text{KHSO}_4 \)[/tex] + 1 from [tex]\( \text{H}_2\text{O} \)[/tex])
- H: 3 (1 from [tex]\( \text{KHSO}_4 \)[/tex] + 2 from [tex]\( \text{H}_2\text{O} \)[/tex])
- S: 1
This equation is balanced.
Option B:
[tex]\[ 2\text{Na} + \text{S} \rightarrow 2\text{NaS} \][/tex]
- On the left side:
- Na: 2
- S: 1
- On the right side:
- Na: 2
- S: 2 (each NaS has 1 S, so 2 NaS has 2 S)
This equation is not balanced because the sulfur atoms are not equal on both sides.
Option C:
[tex]\[ \text{NH}_3 + \text{H}_2\text{O} \rightarrow 2\text{NH}_4\text{OH} \][/tex]
- On the left side:
- N: 1
- H: 5 (3 from [tex]\( \text{NH}_3 \)[/tex] + 2 from [tex]\( \text{H}_2\text{O} \)[/tex])
- O: 1
- On the right side:
- N: 2
- H: 10 (8 from 2 [tex]\( \text{NH}_4 \)[/tex] + 2 from 2 [tex]\( \text{OH} \)[/tex])
- O: 2
This equation is not balanced because the numbers of atoms for each element do not match on both sides.
Option D:
[tex]\[ 2\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{HCl} + \text{NaSO}_4 \][/tex]
- On the left side:
- Na: 2
- Cl: 2
- H: 2
- S: 1
- O: 4
- On the right side:
- Na: 1
- Cl: 1
- H: 1
- S: 1
- O: 4
This equation is not balanced because the numbers of sodium (Na), chlorine (Cl), and hydrogen (H) atoms are not equal on both sides.
From our analysis, the best answer is Option A:
[tex]\[ \text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{KHSO}_4 + \text{H}_2\text{O} \][/tex]
This equation is balanced as it satisfies the condition of having an equal number of each type of atom on both sides of the equation.
