Answer :
To balance the chemical equation [tex]\( N_2O_3 \rightarrow N_2 + O_2 \)[/tex], follow these steps:
1. Balance Nitrogen (N) atoms:
- On the reactant side, [tex]\( N_2O_3 \)[/tex] has 2 Nitrogen atoms.
- On the product side, [tex]\( N_2 \)[/tex] also has 2 Nitrogen atoms.
- Therefore, the Nitrogen atoms are already balanced.
2. Balance Oxygen (O) atoms:
- On the reactant side, [tex]\( N_2O_3 \)[/tex] has 3 Oxygen atoms.
- On the product side, [tex]\( O_2 \)[/tex] has 2 Oxygen atoms per molecule.
- To balance the 3 Oxygen atoms from the reactant side with whole numbers on the product side, we can use fractional coefficients.
However, it is preferable to use whole numbers for balancing:
3. Adjust coefficients to use whole numbers:
- We notice that if we double the reactant, we can balance with whole numbers:
[tex]\( 2 \times N_2O_3 \)[/tex] will give us 6 Oxygen atoms on the reactant side.
- To balance this on the product side, we need 3 [tex]\( O_2 \)[/tex] molecules (since [tex]\( 3 \times 2 = 6 \)[/tex] Oxygen atoms).
4. Final equation:
- We use 2 [tex]\( N_2O_3 \)[/tex] molecules to yield 2 [tex]\( N_2 \)[/tex] molecules and 3 [tex]\( O_2 \)[/tex] molecules.
Hence, the balanced equation is:
[tex]\[ \boxed{2} \, N_2O_3 \rightarrow \boxed{2} \, N_2 + \boxed{3} \, O_2 \][/tex]
1. Balance Nitrogen (N) atoms:
- On the reactant side, [tex]\( N_2O_3 \)[/tex] has 2 Nitrogen atoms.
- On the product side, [tex]\( N_2 \)[/tex] also has 2 Nitrogen atoms.
- Therefore, the Nitrogen atoms are already balanced.
2. Balance Oxygen (O) atoms:
- On the reactant side, [tex]\( N_2O_3 \)[/tex] has 3 Oxygen atoms.
- On the product side, [tex]\( O_2 \)[/tex] has 2 Oxygen atoms per molecule.
- To balance the 3 Oxygen atoms from the reactant side with whole numbers on the product side, we can use fractional coefficients.
However, it is preferable to use whole numbers for balancing:
3. Adjust coefficients to use whole numbers:
- We notice that if we double the reactant, we can balance with whole numbers:
[tex]\( 2 \times N_2O_3 \)[/tex] will give us 6 Oxygen atoms on the reactant side.
- To balance this on the product side, we need 3 [tex]\( O_2 \)[/tex] molecules (since [tex]\( 3 \times 2 = 6 \)[/tex] Oxygen atoms).
4. Final equation:
- We use 2 [tex]\( N_2O_3 \)[/tex] molecules to yield 2 [tex]\( N_2 \)[/tex] molecules and 3 [tex]\( O_2 \)[/tex] molecules.
Hence, the balanced equation is:
[tex]\[ \boxed{2} \, N_2O_3 \rightarrow \boxed{2} \, N_2 + \boxed{3} \, O_2 \][/tex]
