Calculate % yeild of the reaction if 200 g` KHCO_(3)` produces 22 g of `CO_(2)` upon strong heating.

Answer

To calculate the percent yield of the reaction where 200 g of KHCO₃ produces 22 g of CO₂ upon strong heating, we can follow these steps:

### Step 1: Write the balanced chemical equation
The decomposition of potassium bicarbonate (KHCO₃) upon heating can be represented by the following balanced equation:

\[ 2 \text{KHCO}_3 (s) \rightarrow \text{K}_2\text{CO}_3 (s) + \text{CO}_2 (g) + \text{H}_2\text{O} (g) \]

### Step 2: Calculate the moles of KHCO₃ used
The molar mass of KHCO₃ is calculated as follows:
- K: 39.1 g/mol
- H: 1.0 g/mol
- C: 12.0 g/mol
- O: 16.0 g/mol (3 O atoms)

Thus, the molar mass of KHCO₃ is:
\[ 39.1 + 1.0 + 12.0 + (3 \times 16.0) = 100.1 \text{ g/mol} \]

Now, calculate the moles of KHCO₃ used:
\[ \text{Moles of KHCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{200 \text{ g}}{100.1 \text{ g/mol}} \approx 1.99 \text{ moles} \]

### Step 3: Determine the theoretical yield of CO₂
From the balanced equation, we see that 2 moles of KHCO₃ produce 1 mole of CO₂. Thus, the moles of CO₂ produced from 1.99 moles of KHCO₃ is:
\[ \text{Moles of CO}_2 = \frac{1.99 \text{ moles KHCO}_3}{2} \approx 0.995 \text{ moles CO}_2 \]

Now, calculate the mass of CO₂ produced:
The molar mass of CO₂ is:
- C: 12.0 g/mol
- O: 16.0 g/mol (2 O atoms)

Thus, the molar mass of CO₂ is:
\[ 12.0 + (2 \times 16.0) = 44.0 \text{ g/mol} \]

Now, calculate the theoretical yield of CO₂:
\[ \text{Mass of CO}_2 = \text{moles} \times \text{molar mass} = 0.995 \text{ moles} \times 44.0 \text{ g/mol} \approx 43.78 \text{ g} \]

### Step 4: Calculate the percent yield
The percent yield is calculated using the formula:
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]

Substituting the values:
\[ \text{Percent Yield} = \left( \frac{22 \text{ g}}{43.78 \text{ g}} \right) \times 100 \approx 50.3\% \]

### Final Answer
The percent yield of the reaction is approximately **50.3%**.

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