2 moles of `KClO_(3)` is completely decomposed to produce `O_(2)`. How many moles of butane can be completely burnt by the `O_(2)` gas produced

Answer

To solve the problem, we need to follow these steps:

### Step 1: Determine the decomposition of KClO₃
The decomposition reaction of potassium chlorate (KClO₃) is given by the equation:
\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \]
From this balanced equation, we can see that 2 moles of KClO₃ produce 3 moles of O₂.

### Step 2: Calculate the moles of O₂ produced
Since we have 2 moles of KClO₃, we can directly use the stoichiometry from the balanced equation:
\[ \text{Moles of } O_2 = \frac{3 \text{ moles } O_2}{2 \text{ moles } KClO_3} \times 2 \text{ moles } KClO_3 = 3 \text{ moles } O_2 \]

### Step 3: Write the combustion reaction of butane (C₄H₁₀)
The combustion reaction of butane (C₄H₁₀) can be represented as:
\[ \text{C}_4\text{H}_{10} + 6 \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} \]
From this equation, we see that 1 mole of butane requires 6 moles of O₂ for complete combustion.

### Step 4: Calculate how many moles of butane can be burnt with the produced O₂
Now, we have 3 moles of O₂ available for combustion. We can find out how many moles of butane can be completely burnt using the stoichiometry of the combustion reaction:
\[ \text{Moles of } C_4H_{10} = \frac{3 \text{ moles } O_2}{6 \text{ moles } O_2} = 0.5 \text{ moles } C_4H_{10} \]

### Conclusion
Thus, the number of moles of butane that can be completely burnt by the O₂ gas produced from the decomposition of 2 moles of KClO₃ is **0.5 moles**.

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