Question

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

Answer 1

$1$ litre gasoline require $1 \,cc$ of $\left( C _{2} H _{5}\right)_{4} Pb$
$\therefore 10\, \ell=10 \,cc$
$ \Rightarrow 16.6\, g$
$ \Rightarrow \frac{16.6}{322}$ mole
$\therefore$ moles of $C _{2} H _{5} Cl$ veq $=4 \times \frac{16.6}{322}$ moles
$\therefore w _{ C _{2} H _{5} Cl }=4 \times \frac{16.6}{322}(24+3+35.5)=13.3$