Upon heating, baking soda (sodium hydrogen carbonate) decomposes according to the equation:
`2NaHCO_(3)(s) to Na_(2)CO_(3)(s) + H_(2)O(l) + CO_(2)(g)`
In an experiment, 8.4 g baking soda decomposes and carbon dioxide gas is collected.
What would be the volume of `CO_(2)` measured at STP?
`(R.A.M. : Na = 23 , H = 1, C = 12 , O = 16)`

Explanation

To solve the problem of finding the volume of carbon dioxide gas collected at STP from the decomposition of baking soda, we will follow these steps:

### Step 1: Calculate the molar mass of sodium hydrogen carbonate (NaHCO₃).
- Sodium (Na): 23 g/mol
- Hydrogen (H): 1 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol (there are 3 oxygen atoms)

**Molar mass of NaHCO₃ = 23 + 1 + 12 + (16 × 3) = 23 + 1 + 12 + 48 = 84 g/mol**

### Step 2: Calculate the number of moles of NaHCO₃ in 8.4 g.
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
\[ \text{Number of moles of NaHCO₃} = \frac{8.4 \, \text{g}}{84 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 3: Use the stoichiometry of the reaction to find the moles of CO₂ produced.
From the balanced equation:
\[ 2 \, \text{NaHCO₃} \rightarrow \text{Na₂CO₃} + \text{H₂O} + \text{CO₂} \]
This shows that 2 moles of NaHCO₃ produce 1 mole of CO₂. Therefore:
\[ \text{Moles of CO₂} = \frac{0.1 \, \text{mol NaHCO₃}}{2} = 0.05 \, \text{mol CO₂} \]

### Step 4: Calculate the volume of CO₂ at STP.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Thus:
\[ \text{Volume of CO₂} = \text{moles of CO₂} \times 22.4 \, \text{L/mol} \]
\[ \text{Volume of CO₂} = 0.05 \, \text{mol} \times 22.4 \, \text{L/mol} = 1.12 \, \text{L} \]

### Final Answer:
The volume of CO₂ measured at STP is **1.12 liters**.

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