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
Which reactant in limiting reagent (small proportiona)?

Explanation

To determine the limiting reagent in the reaction between methane (\(CH_4\)) and oxygen (\(O_2\)), we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the combustion of methane is:
\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g) + \text{heat} \]

### Step 2: Calculate the number of moles of each reactant
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

- **Molar mass of \(CH_4\)**:
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol × 4 = 4 g/mol
- Total = 12 + 4 = 16 g/mol

- **Molar mass of \(O_2\)**:
- Oxygen (O) = 16 g/mol × 2 = 32 g/mol

Now, calculate the moles:
- Moles of \(CH_4\):
\[ \text{Moles of } CH_4 = \frac{8 \text{ g}}{16 \text{ g/mol}} = 0.5 \text{ moles} \]

- Moles of \(O_2\):
\[ \text{Moles of } O_2 = \frac{24 \text{ g}}{32 \text{ g/mol}} = 0.75 \text{ moles} \]

### Step 3: Determine the stoichiometric requirements
From the balanced equation, we see that 1 mole of \(CH_4\) requires 2 moles of \(O_2\). Therefore, the amount of \(O_2\) needed for 0.5 moles of \(CH_4\) is:
\[ \text{Required moles of } O_2 = 0.5 \text{ moles of } CH_4 \times 2 = 1 \text{ mole of } O_2 \]

### Step 4: Compare the available moles of \(O_2\) with the required moles
We have only 0.75 moles of \(O_2\) available, but we need 1 mole of \(O_2\) to completely react with 0.5 moles of \(CH_4\).

### Step 5: Identify the limiting reagent
Since we do not have enough \(O_2\) to react with all the \(CH_4\), \(O_2\) is the limiting reagent.

### Conclusion
The limiting reagent in this reaction is \(O_2\) because it is present in a smaller proportion than required to completely react with the available \(CH_4\).

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