Questions: [Na2O2 + HOH -> NaOH + O2

Transcript text: $\left[\mathrm{Na}_{2} \mathrm{O}_{2}+\quad \mathrm{HOH} \rightarrow \quad \mathrm{NaOH}+\quad \mathrm{O}_{2}\right.$

Solution

Solution Steps

Step 1: Analyze the chemical equation

The given chemical equation is:

_ Na₂O₂ + _ HOH → _ NaOH + _ O₂

This represents the reaction of sodium peroxide (Na₂O₂) with water (HOH, which is the same as H₂O) to produce sodium hydroxide (NaOH) and oxygen gas (O₂).

Step 2: Balance the sodium (Na) atoms

There are 2 Na atoms on the reactant side and 1 Na atom on the product side. To balance the Na atoms, we put a coefficient of 2 in front of NaOH:

_ Na₂O₂ + _ HOH → 2 NaOH + _ O₂

Step 3: Balance the hydrogen (H) atoms

Now there are 2 H atoms on the reactant side and 2 H atoms on the product side (from the 2 NaOH). So, hydrogen is balanced.

Step 4: Balance the oxygen (O) atoms

There are 2 O atoms from Na₂O₂ and 1 O atom from HOH, totaling 3 O atoms on the reactant side. On the product side, there are 2 O atoms from 2 NaOH molecules, and 2 from the O₂. Thus, there are 4 oxygen atoms on the product side, and 3 oxygen atoms on the reactant side.

To balance the oxygen atoms, we modify the whole equation. Since oxygen is diatomic as O₂, we'll work around even numbers. Since the reactant side has an odd number of oxygens, let's multiply the number of NaOH by 2:

_ Na₂O₂ + _ HOH → 4 NaOH + _ O₂

Now there are 4 oxygens from 4 NaOH. We can't change O₂. To get the sodium to balance again, we multiply the number of Na₂O₂ molecules by 2:

2 Na₂O₂ + _ HOH → 4 NaOH + _ O₂

Now there are 6 oxygen atoms on the reactant side (4 from 2 Na₂O₂ and 2 from HOH or 2H₂O). Since there are 4 oxygen atoms from 4NaOH, the oxygen from O₂ has a coefficient of 1. Then, we must increase the amount of water molecules to balance the oxygen.

2 Na₂O₂ + 2 HOH → 4 NaOH + O₂

Now, there are 6 oxygens on the reactant side and 6 oxygens on the product side.