Answer :
To balance the chemical equation for the reaction between sodium (Na) and oxygen gas (O₂) to form sodium oxide (Na₂O), we need to ensure that the number of atoms for each element is the same on both the reactant and product sides of the equation.
The unbalanced equation is:
[tex]\[ \text{Na}_{(s)} + \text{O}_2{}_{(g)} \rightarrow \text{Na}_2\text{O}_{(s)} \][/tex]
1. Balance Sodium (Na) atoms:
- On the product side, there are 2 sodium atoms in one molecule of \(\text{Na}_2\text{O}\).
- Therefore, we need 2 sodium atoms on the reactant side to balance the sodium atoms.
[tex]\[ 2\text{Na}_{(s)} + \text{O}_2{}_{(g)} \rightarrow \text{Na}_2\text{O}_{(s)} \][/tex]
2. Balance Oxygen (O) atoms:
- On the reactant side, we have \(\text{O}_2\) which contains 2 oxygen atoms.
- On the product side, \(\text{Na}_2\text{O}\) contains 1 oxygen atom per molecule.
- To balance the oxygen atoms, we need two molecules of \(\text{Na}_2\text{O}\) on the product side, which will give us a total of 2 oxygen atoms.
[tex]\[ 2\text{Na}_{(s)} + \text{O}_2{}_{(g)} \rightarrow 2\text{Na}_2\text{O}_{(s)} \][/tex]
3. Adjust the Sodium (Na) atoms again:
- Now, by having 2 molecules of \(\text{Na}_2\text{O}\) on the product side, we need 4 sodium atoms on the reactant side.
[tex]\[ 4\text{Na}_{(s)} + \text{O}_2{}_{(g)} \rightarrow 2\text{Na}_2\text{O}_{(s)} \][/tex]
Hence, the balanced chemical equation is:
[tex]\[ 4\text{Na}_{(s)} + \text{O}_2{}_{(g)} \rightarrow 2\text{Na}_2\text{O}_{(s)} \][/tex]
This equation maintains an equal number of sodium (Na) and oxygen (O) atoms on both sides of the reaction.
The unbalanced equation is:
[tex]\[ \text{Na}_{(s)} + \text{O}_2{}_{(g)} \rightarrow \text{Na}_2\text{O}_{(s)} \][/tex]
1. Balance Sodium (Na) atoms:
- On the product side, there are 2 sodium atoms in one molecule of \(\text{Na}_2\text{O}\).
- Therefore, we need 2 sodium atoms on the reactant side to balance the sodium atoms.
[tex]\[ 2\text{Na}_{(s)} + \text{O}_2{}_{(g)} \rightarrow \text{Na}_2\text{O}_{(s)} \][/tex]
2. Balance Oxygen (O) atoms:
- On the reactant side, we have \(\text{O}_2\) which contains 2 oxygen atoms.
- On the product side, \(\text{Na}_2\text{O}\) contains 1 oxygen atom per molecule.
- To balance the oxygen atoms, we need two molecules of \(\text{Na}_2\text{O}\) on the product side, which will give us a total of 2 oxygen atoms.
[tex]\[ 2\text{Na}_{(s)} + \text{O}_2{}_{(g)} \rightarrow 2\text{Na}_2\text{O}_{(s)} \][/tex]
3. Adjust the Sodium (Na) atoms again:
- Now, by having 2 molecules of \(\text{Na}_2\text{O}\) on the product side, we need 4 sodium atoms on the reactant side.
[tex]\[ 4\text{Na}_{(s)} + \text{O}_2{}_{(g)} \rightarrow 2\text{Na}_2\text{O}_{(s)} \][/tex]
Hence, the balanced chemical equation is:
[tex]\[ 4\text{Na}_{(s)} + \text{O}_2{}_{(g)} \rightarrow 2\text{Na}_2\text{O}_{(s)} \][/tex]
This equation maintains an equal number of sodium (Na) and oxygen (O) atoms on both sides of the reaction.
