A particular 100 -octane aviation gasoline used 1 cc of ( C 2 H 5)4 Pb, of density 1.66 gm / cc, per litre of gasoline. ( C 2 H 5)4 Pb is made as follows: 4 C 2 H 6 Cl +4 NaPb longrightarrow( C 2 H 5)4 Pb +4 NaCl How many gram of C 2 H 5 Cl is needed to make enough ( C 2 H 5)4 Pb for 10 litre of gasoline. (atomic mass of Pb =206 )

Question

A particular 100 -octane aviation gasoline used 1 cc of ( C 2 H 5)4 Pb, of density 1.66 gm / cc, per litre of gasoline. ( C 2 H 5)4 Pb is made as follows: 4 C 2 H 6 Cl +4 NaPb longrightarrow( C 2 H 5)4 Pb +4 NaCl How many gram of C 2 H 5 Cl is needed to make enough ( C 2 H 5)4 Pb for 10 litre of gasoline. (atomic mass of Pb =206 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

1 litre gasoline require 1 cc of ( C 2 H 5)4 Pb ∴ 10 ℓ=10 cc ⇒ 16.6 g ⇒ (16.6/322) mole ∴ moles of C 2 H 5 Cl veq =4 × (16.6/322) moles ∴ w C 2 H 5 Cl =4 × (16.6/322)(24+3+35.5)=13.3

1 litre gasoline require 1cc of (C2​H5​)4​Pb ∴10ℓ=10cc ⇒16.6g ⇒32216.6​ mole ∴ moles of C2​H5​Cl veq =4×32216.6​ moles ∴wC2​H5​Cl​=4×32216.6​(24+3+35.5)=13.3

$1$ litre gasoline require $1 cc$ of $(C_{2} H_{5})_{4} P b$
$∴ 10 ℓ = 10 cc$
$\Rightarrow 16.6 g$
$\Rightarrow \frac{16.6}{322}$ mole
$∴$ moles of $C_{2} H_{5} Cl$ veq $= 4 \times \frac{16.6}{322}$ moles
$∴ w_{C_{2} H_{5} Cl} = 4 \times \frac{16.6}{322} (24 + 3 + 35.5) = 13.3$