In an experiment, a mixture of 8g `CH_(4) and 24gO_(2)` was burn to form `CO_(2) and H_(2)O` according to the equation `:CH_(4)(g)+2O_(2)(g) to CO_(2)(g)+2H_(2)O(g)+` heat
How many moles of `CO_(2)` will be formed ?

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the combustion of methane (CH₄) is:
\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g) + \text{heat} \]

### Step 2: Calculate the number of moles of CH₄ and O₂
- **Molar mass of CH₄ (Methane)**:
\[ \text{C} = 12 \, \text{g/mol}, \, \text{H} = 1 \, \text{g/mol} \]
\[ \text{Molar mass of CH}_4 = 12 + (4 \times 1) = 16 \, \text{g/mol} \]

- **Molar mass of O₂ (Oxygen)**:
\[ \text{O} = 16 \, \text{g/mol} \]
\[ \text{Molar mass of O}_2 = 2 \times 16 = 32 \, \text{g/mol} \]

- **Moles of CH₄**:
\[ \text{Moles of CH}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{8 \, \text{g}}{16 \, \text{g/mol}} = 0.5 \, \text{moles} \]

- **Moles of O₂**:
\[ \text{Moles of O}_2 = \frac{24 \, \text{g}}{32 \, \text{g/mol}} = 0.75 \, \text{moles} \]

### Step 3: Determine the limiting reactant
From the balanced equation, 1 mole of CH₄ reacts with 2 moles of O₂. Therefore, for 0.5 moles of CH₄, the required moles of O₂ are:
\[ 0.5 \, \text{moles of CH}_4 \times 2 \, \text{moles of O}_2/\text{mole of CH}_4 = 1 \, \text{mole of O}_2 \]

Since we only have 0.75 moles of O₂ available, which is less than the required 1 mole, O₂ is the limiting reactant.

### Step 4: Calculate the moles of CO₂ produced
According to the stoichiometry of the reaction:
- 2 moles of O₂ produce 1 mole of CO₂.

Thus, the moles of CO₂ produced from the available O₂ (0.75 moles) is:
\[ \text{Moles of CO}_2 = \frac{0.75 \, \text{moles of O}_2}{2} = 0.375 \, \text{moles of CO}_2 \]

### Conclusion
The number of moles of CO₂ formed from the combustion of 8 g of CH₄ and 24 g of O₂ is **0.375 moles**.

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