The volume of `CO_(2)` liberated at STP is _________ , on thermal decomposition of 84 g of sodium bicarbonate.

Answer

To solve the problem of finding the volume of CO₂ liberated at STP from the thermal decomposition of 84 g of sodium bicarbonate (NaHCO₃), we can follow these steps:

### Step 1: Write the balanced chemical equation for the thermal decomposition of sodium bicarbonate.

The thermal decomposition of sodium bicarbonate can be represented by the following balanced equation:

\[ 2 \text{NaHCO}_3 (s) \rightarrow \text{Na}_2\text{CO}_3 (s) + \text{CO}_2 (g) + \text{H}_2\text{O} (l) \]

### Step 2: Calculate the molar mass of sodium bicarbonate (NaHCO₃).

To find the molar mass, we add the atomic masses of each element in the compound:

- Sodium (Na): 23 g/mol
- Hydrogen (H): 1 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol

Now, sum these values:

\[ \text{Molar mass of NaHCO}_3 = 23 + 1 + 12 + 48 = 84 \text{ g/mol} \]

### Step 3: Calculate the number of moles of sodium bicarbonate in 84 g.

Using the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Substituting the values:

\[ \text{Number of moles of NaHCO}_3 = \frac{84 \text{ g}}{84 \text{ g/mol}} = 1 \text{ mole} \]

### Step 4: Use the stoichiometry of the reaction to find the moles of CO₂ produced.

From the balanced equation, we see that:

\[ 2 \text{ moles of NaHCO}_3 \rightarrow 1 \text{ mole of CO}_2 \]

Thus, for 1 mole of NaHCO₃, the moles of CO₂ produced will be:

\[ \text{Moles of CO}_2 = \frac{1 \text{ mole NaHCO}_3}{2} = 0.5 \text{ moles of CO}_2 \]

### Step 5: Calculate the volume of CO₂ at STP.

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of 0.5 moles of CO₂ can be calculated as follows:

\[ \text{Volume of CO}_2 = 0.5 \text{ moles} \times 22.4 \text{ L/mole} = 11.2 \text{ L} \]

### Final Answer:

The volume of CO₂ liberated at STP is **11.2 liters**.

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