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 other reactant will remain unreacted after the reaction has stopped ?

Explanation

To solve the problem, 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 molar masses
- Molar mass of CH₄ (methane) = 12 (C) + 4 (H) = 16 g/mol
- Molar mass of O₂ (oxygen) = 32 g/mol

### Step 3: Calculate the number of moles of each reactant
- 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 4: Determine the limiting reactant
From the balanced equation, we see that 1 mole of CH₄ reacts with 2 moles of O₂. Therefore:
- To completely react with 0.5 moles of CH₄, we need:
\[ 0.5 \text{ moles CH}_4 \times 2 = 1 \text{ mole O}_2 \]

Since we only have 0.75 moles of O₂, O₂ is the limiting reactant.

### Step 5: Calculate the amount of CH₄ that reacts with the available O₂
From the balanced equation:
- 2 moles of O₂ react with 1 mole of CH₄.
- Therefore, 0.75 moles of O₂ will react with:
\[ \text{Moles of CH}_4 = \frac{0.75}{2} = 0.375 \text{ moles} \]

### Step 6: Calculate the unreacted CH₄
Initially, we had 0.5 moles of CH₄. The amount that reacted is 0.375 moles. Thus, the unreacted CH₄ is:
\[ \text{Unreacted CH}_4 = 0.5 - 0.375 = 0.125 \text{ moles} \]

### Final Answer
0.125 moles of CH₄ remains unreacted after the reaction has stopped.

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