Give balance equation for the reaction of aluminium with aqueous sodium hydroxide.

Answer

To write the balanced equation for the reaction of aluminium with aqueous sodium hydroxide, we will follow these steps:

### Step 1: Write the unbalanced equation
The initial reaction involves aluminium (Al) and sodium hydroxide (NaOH) in aqueous form. We also need to consider water (H2O) as a reactant since the reaction occurs in an aqueous solution.

**Unbalanced Equation:**
\[ \text{Al} + \text{NaOH} + \text{H}_2\text{O} \rightarrow \text{NaAlO}_2 + \text{H}_2 \]

### Step 2: Identify the products
When aluminium reacts with sodium hydroxide and water, it forms sodium aluminate (NaAlO2) and hydrogen gas (H2).

### Step 3: Balance the equation
Now we will balance the equation. We need to ensure that the number of atoms of each element is the same on both sides of the equation.

1. Start with aluminium (Al). There is 1 Al on the left and 1 Al in NaAlO2 on the right.
2. Next, balance sodium (Na). There is 1 Na in NaOH and 1 Na in NaAlO2, so they are balanced.
3. Now, balance oxygen (O) and hydrogen (H).

The unbalanced equation is:
\[ \text{Al} + \text{NaOH} + \text{H}_2\text{O} \rightarrow \text{NaAlO}_2 + \text{H}_2 \]

To balance it, we can multiply the reactants and products by appropriate coefficients:
- 2 Al on the left will give us 2 NaAlO2 on the right.
- 2 NaOH and 2 water molecules will provide the necessary sodium and hydroxide ions.

The balanced equation becomes:
\[ 2 \text{Al} + 2 \text{NaOH} + 2 \text{H}_2\text{O} \rightarrow 2 \text{NaAlO}_2 + 3 \text{H}_2 \]

### Final Balanced Equation:
\[ 2 \text{Al} + 2 \text{NaOH} + 2 \text{H}_2\text{O} \rightarrow 2 \text{NaAlO}_2 + 3 \text{H}_2 \]

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