When a sample of baking is strongly ignited in a crucible, it suffered a loss in weight of 3.1 g. The mass of baking soda is

Options

A. a. 16.8 g
B. b. 8.4 g
C. c. 11.6 g
D. d. 4.2 g

Answer

To solve the problem of finding the mass of baking soda (sodium bicarbonate, NaHCO₃) that suffered a loss in weight of 3.1 g upon ignition, we can follow these steps:

### Step 1: Write the balanced chemical equation
The decomposition of sodium bicarbonate upon heating can be represented by the following balanced equation:

\[ 2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \]

### Step 2: Determine the molar masses
We need to find the molar masses of the relevant compounds:
- Molar mass of NaHCO₃ = 23 (Na) + 1 (H) + 12 (C) + 3 × 16 (O) = 84 g/mol
- Molar mass of CO₂ = 12 (C) + 2 × 16 (O) = 44 g/mol

### Step 3: Relate the loss of mass to the moles of CO₂ produced
From the balanced equation, we see that 2 moles of NaHCO₃ produce 1 mole of CO₂. This means that the loss of mass (which is due to the release of CO₂ and H₂O) can be calculated based on the molar masses.

### Step 4: Set up the proportion
The mass loss corresponds to the moles of CO₂ produced. Since 2 moles of NaHCO₃ produce 1 mole of CO₂, we can set up the following proportion based on the molar masses:

\[ \text{Loss in mass (3.1 g)} = \frac{1 \text{ mole CO}_2}{2 \text{ moles NaHCO}_3} \times \text{mass of NaHCO}_3 \]

### Step 5: Calculate the mass of NaHCO₃
Using the proportion, we can express the mass of NaHCO₃ in terms of the mass loss:

\[ x = \frac{3.1 \, \text{g} \times 2 \times 84 \, \text{g/mol}}{44 \, \text{g/mol}} \]

### Step 6: Perform the calculation
Now, we can calculate the mass of NaHCO₃:

\[ x = \frac{3.1 \times 2 \times 84}{44} \]
\[ x = \frac{520.8}{44} \]
\[ x \approx 11.85 \, \text{g} \]

### Final Answer
The mass of baking soda (NaHCO₃) is approximately **11.85 g**.

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