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 grams of `CO_(2)` will be formed ?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation.
The balanced 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 of the reactants.
- Molar mass of CH₄ (methane) = 12 (C) + 4 (H) = 16 g/mol
- Molar mass of O₂ (oxygen) = 2 × 16 = 32 g/mol

### Step 3: Calculate the number of moles of each reactant.
- Moles of CH₄ = \(\frac{8 \text{ g}}{16 \text{ g/mol}} = 0.5 \text{ moles}\)
- Moles of O₂ = \(\frac{24 \text{ g}}{32 \text{ g/mol}} = 0.75 \text{ moles}\)

### Step 4: Determine the limiting reagent.
According to 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:
\[ 0.5 \text{ moles CH}_4 \times 2 \text{ moles O}_2/\text{mole CH}_4 = 1 \text{ mole O}_2 \]
Since we only have 0.75 moles of O₂ available, O₂ is the limiting reagent.

### Step 5: Calculate the amount of CO₂ produced.
From the balanced equation, 2 moles of O₂ produce 1 mole of CO₂. Therefore, 0.75 moles of O₂ will produce:
\[ \frac{0.75 \text{ moles O}_2}{2} = 0.375 \text{ moles CO}_2 \]

### Step 6: Convert moles of CO₂ to grams.
The molar mass of CO₂ is:
\[ 12 (C) + 2 \times 16 (O) = 44 \text{ g/mol} \]
Now, convert moles of CO₂ to grams:
\[ 0.375 \text{ moles CO}_2 \times 44 \text{ g/mol} = 16.5 \text{ g CO}_2 \]

### Final Answer:
The mass of CO₂ produced is **16.5 grams**.

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