Question

In aviation gasoline of $100$ octane number, $1.0$ mL of tetraethyl lead (TEL), $(C_2{H}_5{)}_{4}Pb$ of density $1.615$ g/mL per litre is added to the product. TEL is prepared as follows:
$4C_2{H}_{5}Cl+4Na\left(Pb\right)\to (C_2{H}_5{)}_{4}Pb+4NaCl+3Pb$
Calculate the amount of $C_2{H}_{5}Cl$ required to make enough TEL for $1.0L$ of gasoline.

• A. $0.645$ g
• B. $1.29$ g
• C. $1.935$ g
• D. $2.58$ g

Answer 1

The correct option is B $1.29$ g

The weight of $1.0$ mL TEL $=10\times 1.615=1.615$ g

The weight of TEL needed per litre of gasoline $=1.615$ g
Molecular weight of TEL, $(C_2{H}_5{)}_{4}Pb=4\times 29+207=323$ g/mol
Moles of TEL $=\frac{1.615}{323}$ g/mol $=0.005$ mol
$1$ mol of TEL requires $=4$ mol of $C_2{H}_{5}Cl$
$0.005$ mol of TEL requires $=4\times 0.005$ $=0.02$ mol of $C_2{H}_{5}Cl$
Weight of $C_2{H}_{5}Cl=0.02\times 4.5=1.2$ g