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To balance the chemical equation [tex]\( N_2O_3 \rightarrow N_2 + O_2 \)[/tex], we need to make sure the number of atoms of each element is the same on both the reactant and product sides of the equation. Here is the step-by-step process to balance this equation:
1. Write down the initial equation:
[tex]\( N_2O_3 \rightarrow N_2 + O_2 \)[/tex]
2. Count the number of each type of atom on both sides:
- On the reactant side:
- Nitrogen (N) atoms: 2
- Oxygen (O) atoms: 3
- On the product side:
- Nitrogen (N) atoms: 2 (from [tex]\( N_2 \)[/tex])
- Oxygen (O) atoms: 2 (from [tex]\( O_2 \)[/tex])
3. Balance Oxygen atoms:
The reactant has 3 oxygen atoms while the product has 2. To balance the equation, we need whole molecules. This equation gets straightforward if we use fractional coefficients initially and then clear them by multiplication.
We can use 1.5 [tex]\( O_2 \)[/tex] molecules to balance:
[tex]\( N_2O_3 \rightarrow N_2 + 1.5 O_2 \)[/tex]
4. Convert to whole numbers:
Since chemical equations typically use whole numbers for coefficients, we can multiply the entire equation by 2 to eliminate the fraction.
[tex]\( 2 N_2O_3 \rightarrow 2 N_2 + 3 O_2 \)[/tex]
5. Verify the balance:
- On the left side:
- Nitrogen (N) atoms: 4 (from 2 [tex]\( N_2O_3 \)[/tex])
- Oxygen (O) atoms: 6 (from 2 [tex]\( N_2O_3 \)[/tex])
- On the right side:
- Nitrogen (N) atoms: 4 (from 2 [tex]\( N_2 \)[/tex])
- Oxygen (O) atoms: 6 (from 3 [tex]\( O_2 \)[/tex])
Both sides are balanced, thus the correct coefficients are:
[tex]\[ 2 N_2O_3 \rightarrow 2 N_2 + 3 O_2 \][/tex]
So the balanced chemical equation is:
[tex]\[ \boxed{2} \ N_2O_3 \rightarrow \boxed{2} \ N_2 + \boxed{3} \ O_2 \][/tex]
1. Write down the initial equation:
[tex]\( N_2O_3 \rightarrow N_2 + O_2 \)[/tex]
2. Count the number of each type of atom on both sides:
- On the reactant side:
- Nitrogen (N) atoms: 2
- Oxygen (O) atoms: 3
- On the product side:
- Nitrogen (N) atoms: 2 (from [tex]\( N_2 \)[/tex])
- Oxygen (O) atoms: 2 (from [tex]\( O_2 \)[/tex])
3. Balance Oxygen atoms:
The reactant has 3 oxygen atoms while the product has 2. To balance the equation, we need whole molecules. This equation gets straightforward if we use fractional coefficients initially and then clear them by multiplication.
We can use 1.5 [tex]\( O_2 \)[/tex] molecules to balance:
[tex]\( N_2O_3 \rightarrow N_2 + 1.5 O_2 \)[/tex]
4. Convert to whole numbers:
Since chemical equations typically use whole numbers for coefficients, we can multiply the entire equation by 2 to eliminate the fraction.
[tex]\( 2 N_2O_3 \rightarrow 2 N_2 + 3 O_2 \)[/tex]
5. Verify the balance:
- On the left side:
- Nitrogen (N) atoms: 4 (from 2 [tex]\( N_2O_3 \)[/tex])
- Oxygen (O) atoms: 6 (from 2 [tex]\( N_2O_3 \)[/tex])
- On the right side:
- Nitrogen (N) atoms: 4 (from 2 [tex]\( N_2 \)[/tex])
- Oxygen (O) atoms: 6 (from 3 [tex]\( O_2 \)[/tex])
Both sides are balanced, thus the correct coefficients are:
[tex]\[ 2 N_2O_3 \rightarrow 2 N_2 + 3 O_2 \][/tex]
So the balanced chemical equation is:
[tex]\[ \boxed{2} \ N_2O_3 \rightarrow \boxed{2} \ N_2 + \boxed{3} \ O_2 \][/tex]
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