How much potassium chlorate is needed to get enough oxygen for completely burning 28 g of carbon monoxide ?

Answer

To determine how much potassium chlorate (KClO3) is needed to produce enough oxygen to completely burn 28 g of carbon monoxide (CO), we can follow these steps:

### Step 1: Calculate the number of moles of carbon monoxide (CO).
The number of moles can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
Given mass of CO = 28 g
Molar mass of CO = 28 g/mol
\[ \text{Number of moles of CO} = \frac{28 \text{ g}}{28 \text{ g/mol}} = 1 \text{ mole} \]

### Step 2: Write the balanced chemical equation for the combustion of carbon monoxide.
The combustion of carbon monoxide in the presence of oxygen is represented as:
\[ 2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2 \]
From the equation, we see that 1 mole of CO requires 0.5 moles of O2 for complete combustion.

### Step 3: Determine the amount of oxygen required for 1 mole of CO.
From the balanced equation, we know:
\[ 1 \text{ mole of CO} \text{ requires } 0.5 \text{ moles of O}_2 \]

### Step 4: Calculate the moles of KClO3 needed to produce the required oxygen.
The decomposition of potassium chlorate (KClO3) is represented as:
\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \]
From this equation, we see that:
- 2 moles of KClO3 produce 3 moles of O2.
- Therefore, 1 mole of O2 requires \(\frac{2}{3}\) moles of KClO3.

Now, since we need 0.5 moles of O2:
\[ \text{Moles of KClO}_3 = \frac{2}{3} \times 0.5 = \frac{1}{3} \text{ moles} \]

### Step 5: Calculate the mass of KClO3 required.
To find the mass, we use the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
Molar mass of KClO3 = 122.55 g/mol
\[ \text{Mass of KClO}_3 = \frac{1}{3} \text{ moles} \times 122.55 \text{ g/mol} \approx 40.85 \text{ g} \]

### Final Answer:
Approximately **40.85 grams** of potassium chlorate (KClO3) is needed to produce enough oxygen to completely burn 28 g of carbon monoxide (CO).

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