In aviation gasoline of 100 octane number, `1.0 mL` of tetraethy lead `(TEL), (C_(2)H_(5))_(4) Pb`, of density `1.615 g mL^(-1)`, per litre is added to the product. `TEL` is prepared as follows:
`4C_(2) H_(5) Cl + 4Na(Pb) rarr (C_(2)H_(5))_(4) Pb + 4NaCl + 3 Pb`
Calculate the amount of `C_(2) H_(5) Cl` required to make enough `TEL` for `1.0 L` of gasoline.

Options

A. `0.645 g`
B. ` 1.29 g`
C. `1.935 g`
D. `2.58 g`

Explanation

To solve the problem, we need to calculate the amount of C₂H₅Cl required to prepare 1.0 mL of tetraethyl lead (TEL) for aviation gasoline. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the mass of TEL in 1.0 mL
We know the density of TEL is 1.615 g/mL. The mass can be calculated using the formula:
\[ \text{Mass} = \text{Density} \times \text{Volume} \]
Substituting the values:
\[ \text{Mass of TEL} = 1.615 \, \text{g/mL} \times 1.0 \, \text{mL} = 1.615 \, \text{g} \]

### Step 2: Calculate the molar mass of TEL
The molecular formula of TEL is (C₂H₅)₄Pb. We calculate the molar mass as follows:
- Carbon (C): 12.01 g/mol × 8 (from 4 C₂H₅) = 96.08 g/mol
- Hydrogen (H): 1.008 g/mol × 20 (from 4 C₂H₅) = 20.16 g/mol
- Lead (Pb): 207.2 g/mol × 1 = 207.2 g/mol

Adding these together:
\[ \text{Molar mass of TEL} = 96.08 + 20.16 + 207.2 = 323.44 \, \text{g/mol} \approx 323 \, \text{g/mol} \]

### Step 3: Calculate the number of moles of TEL
Using the mass calculated in Step 1 and the molar mass from Step 2, we can find the number of moles of TEL:
\[ \text{Number of moles of TEL} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1.615 \, \text{g}}{323 \, \text{g/mol}} \approx 0.005 \, \text{mol} \]

### Step 4: Determine the number of moles of C₂H₅Cl required
From the reaction provided, 1 mole of TEL requires 4 moles of C₂H₅Cl. Therefore, the number of moles of C₂H₅Cl required is:
\[ \text{Number of moles of C₂H₅Cl} = 4 \times \text{Number of moles of TEL} = 4 \times 0.005 \, \text{mol} = 0.02 \, \text{mol} \]

### Step 5: Calculate the molar mass of C₂H₅Cl
Next, we calculate the molar mass of C₂H₅Cl:
- Carbon (C): 12.01 g/mol × 2 = 24.02 g/mol
- Hydrogen (H): 1.008 g/mol × 5 = 5.04 g/mol
- Chlorine (Cl): 35.45 g/mol × 1 = 35.45 g/mol

Adding these together:
\[ \text{Molar mass of C₂H₅Cl} = 24.02 + 5.04 + 35.45 = 64.51 \, \text{g/mol} \approx 64.5 \, \text{g/mol} \]

### Step 6: Calculate the mass of C₂H₅Cl required
Finally, we can find the mass of C₂H₅Cl required using the number of moles calculated in Step 4:
\[ \text{Mass of C₂H₅Cl} = \text{Number of moles} \times \text{Molar mass} = 0.02 \, \text{mol} \times 64.5 \, \text{g/mol} = 1.29 \, \text{g} \]

### Final Answer
The amount of C₂H₅Cl required to make enough TEL for 1.0 L of gasoline is **1.29 grams**.

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