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`

Answer

B

Explanation

The weight of `1.0 mL Tel = 1.0 mL xx 1.615 = 1.615 g`
The weight of `TEL` needed per litre of gasoline `= 1.615 g`
`[Mw "of" TEL [C_(2) H_(5))_(4) Pb] = 4 xx 29 + 207 = 323 g mol^(-1))`
moles of `TEL = (1.615)/(323 g mol^(-1)) 0.005 "mol"`
1 mol of `TEL` resquires `= 4 xx 0.005`
`= 0.02 "mo of" C_(2)H_(5) Cl`
Weight of `C_(2)H_(5) Cl = 0.02 xx (64.5 g//"mol") = 1.29 g`