280 g of a mixture containing `CH_(4) and C_(2)H_(6)` in `5:2` molar ratio is burnt in presence of excess of oxygen. Calculate total moles of `CO_(2)` produced.

Answer

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

### Step 1: Determine the mass of each component in the mixture
Given that the molar ratio of `CH₄` to `C₂H₆` is `5:2`, we can denote the mass of `CH₄` as \( x \) grams and the mass of `C₂H₆` as \( 280 - x \) grams.

### Step 2: Set up the molar ratio equation
Using the molar masses:
- Molar mass of `CH₄` = 16 g/mol
- Molar mass of `C₂H₆` = 30 g/mol

The number of moles of `CH₄` and `C₂H₆` can be expressed as:
- Moles of `CH₄` = \( \frac{x}{16} \)
- Moles of `C₂H₆` = \( \frac{280 - x}{30} \)

According to the given molar ratio:
\[ \frac{\frac{x}{16}}{\frac{280 - x}{30}} = \frac{5}{2} \]

### Step 3: Cross-multiply to solve for \( x \)
Cross-multiplying gives:
\[ 2x \cdot 30 = 5(280 - x) \cdot 16 \]
\[ 60x = 5(4480 - 16x) \]
\[ 60x = 22400 - 80x \]
\[ 60x + 80x = 22400 \]
\[ 140x = 22400 \]
\[ x = \frac{22400}{140} = 160 \text{ g} \]

### Step 4: Calculate the mass of `C₂H₆`
Now, substituting \( x \) back to find the mass of `C₂H₆`:
\[ \text{Mass of } C₂H₆ = 280 - x = 280 - 160 = 120 \text{ g} \]

### Step 5: Calculate the number of moles of each component
Now we can find the moles of `CH₄` and `C₂H₆`:
- Moles of `CH₄` = \( \frac{160}{16} = 10 \text{ moles} \)
- Moles of `C₂H₆` = \( \frac{120}{30} = 4 \text{ moles} \)

### Step 6: Write the balanced combustion reactions
1. Combustion of `CH₄`:
\[ CH₄ + 2O₂ \rightarrow CO₂ + 2H₂O \]
From this reaction, 1 mole of `CH₄` produces 1 mole of `CO₂`.

2. Combustion of `C₂H₆`:
\[ C₂H₆ + \frac{7}{2}O₂ \rightarrow 2CO₂ + 3H₂O \]
From this reaction, 1 mole of `C₂H₆` produces 2 moles of `CO₂`.

### Step 7: Calculate total moles of `CO₂` produced
- From `CH₄`: \( 10 \text{ moles} \times 1 = 10 \text{ moles of } CO₂ \)
- From `C₂H₆`: \( 4 \text{ moles} \times 2 = 8 \text{ moles of } CO₂ \)

### Step 8: Total moles of `CO₂`
Total moles of `CO₂` produced:
\[ 10 + 8 = 18 \text{ moles of } CO₂ \]

### Final Answer
The total moles of `CO₂` produced is **18 moles**.

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