Answer :
Certainly! Let's balance the given chemical equation:
[tex]\[ \text{KClO}_3 \rightarrow \text{KCl} + \text{O}_2 \][/tex]
We need to ensure that the number of atoms for each element on the reactant side equals the number of atoms on the product side.
### Step-by-Step Solution:
1. Write the Unbalanced Equation:
[tex]\[ \text{KClO}_3 \rightarrow \text{KCl} + \text{O}_2 \][/tex]
2. Identify the Number of Atoms of Each Element:
- Potassium (K): 1 atom on reactant side (KClO[tex]\(_3\)[/tex]), 1 atom on product side (KCl)
- Chlorine (Cl): 1 atom on reactant side (KClO[tex]\(_3\)[/tex]), 1 atom on product side (KCl)
- Oxygen (O): 3 atoms on reactant side (KClO[tex]\(_3\)[/tex]), 2 atoms on product side (O[tex]\(_2\)[/tex])
3. Balance the Oxygen Atoms:
- To balance the oxygen atoms, we need to make the number of oxygen atoms in the products equal to 6 (using the least common multiple of 3 and 2). Since O[tex]\(_2\)[/tex] molecules have 2 oxygen atoms, we place a coefficient of 3 before O[tex]\(_2\)[/tex]:
[tex]\[ \text{KClO}_3 \rightarrow \text{KCl} + 3/2 \text{O}_2 \][/tex]
- To avoid fractions, multiply the entire equation by 2:
[tex]\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \][/tex]
4. Verify the Balance:
- Potassium (K): 2 atoms on both reactant (2 KClO[tex]\(_3\)[/tex]) and product (2 KCl) sides
- Chlorine (Cl): 2 atoms on both reactant (2 KClO[tex]\(_3\)[/tex]) and product (2 KCl) sides
- Oxygen (O): 6 atoms on both reactant (2 KClO[tex]\(_3\)[/tex]) and product (3 O[tex]\(_2\)[/tex]) sides
Thus, the number of atoms for each element is balanced on both sides of the equation.
### Balanced Equation:
[tex]\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \][/tex]
The correct coefficients for the balanced equation are:
- [tex]\( \text{KClO}_3 : 2 \)[/tex]
- [tex]\( \text{KCl} : 2 \)[/tex]
- [tex]\( \text{O}_2 : 3 \)[/tex]
[tex]\[ \text{KClO}_3 \rightarrow \text{KCl} + \text{O}_2 \][/tex]
We need to ensure that the number of atoms for each element on the reactant side equals the number of atoms on the product side.
### Step-by-Step Solution:
1. Write the Unbalanced Equation:
[tex]\[ \text{KClO}_3 \rightarrow \text{KCl} + \text{O}_2 \][/tex]
2. Identify the Number of Atoms of Each Element:
- Potassium (K): 1 atom on reactant side (KClO[tex]\(_3\)[/tex]), 1 atom on product side (KCl)
- Chlorine (Cl): 1 atom on reactant side (KClO[tex]\(_3\)[/tex]), 1 atom on product side (KCl)
- Oxygen (O): 3 atoms on reactant side (KClO[tex]\(_3\)[/tex]), 2 atoms on product side (O[tex]\(_2\)[/tex])
3. Balance the Oxygen Atoms:
- To balance the oxygen atoms, we need to make the number of oxygen atoms in the products equal to 6 (using the least common multiple of 3 and 2). Since O[tex]\(_2\)[/tex] molecules have 2 oxygen atoms, we place a coefficient of 3 before O[tex]\(_2\)[/tex]:
[tex]\[ \text{KClO}_3 \rightarrow \text{KCl} + 3/2 \text{O}_2 \][/tex]
- To avoid fractions, multiply the entire equation by 2:
[tex]\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \][/tex]
4. Verify the Balance:
- Potassium (K): 2 atoms on both reactant (2 KClO[tex]\(_3\)[/tex]) and product (2 KCl) sides
- Chlorine (Cl): 2 atoms on both reactant (2 KClO[tex]\(_3\)[/tex]) and product (2 KCl) sides
- Oxygen (O): 6 atoms on both reactant (2 KClO[tex]\(_3\)[/tex]) and product (3 O[tex]\(_2\)[/tex]) sides
Thus, the number of atoms for each element is balanced on both sides of the equation.
### Balanced Equation:
[tex]\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \][/tex]
The correct coefficients for the balanced equation are:
- [tex]\( \text{KClO}_3 : 2 \)[/tex]
- [tex]\( \text{KCl} : 2 \)[/tex]
- [tex]\( \text{O}_2 : 3 \)[/tex]
