Question

A natural gas sample contains $84$ % (by volume) of $C{H}_4,10$ % of $C_2{H}_6,3$ % of $C_3{H}_8$ and $3$ % of $N_2$ . If a series of catalytic reactions could be used for converting all the carbon atoms of the gas in to butadiene , $C_4{H}_6$ , with $100$ % efficiency , how much butadiene could be prepared from $100g$ of the natural gas ?

Answer 1

$W(C{H}_4+C_2{H}_6+C_3{H}_7+N_2)100g$

$\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$

% by volume $84$ % $10$ % $3$ % $3$ %

We know , volume % + mole % ( at constant temperature and pressure )

mole % of $C{H}_4=84$ % & mole % of $C_2{H}_6=10$ %

mole % of $C_3{H}_8=3$ % & mole of $N_2=3$ %

By POAC

mole of $C$ before reaction = mole of $C$ after reaction

mole of $C$ ( in $C{H}_4$ + in $C_2{H}_6$ + in $C_3{H}_8$ ) = mole of $C$ in $C_4{H}_6$

mole of $C{H}_4+2$ mole of $C_2{H}_6+3\times$ mole of $C_3{H}_8=4\times$ mole of $C_{4}H-6$

$84+2\times 10+3\times 3=4\times$ mole of $C_4{H}_6$

$84+20+9=4\times$ mole of $C_4{H}_6$

mole of $C_4{H}_6=\frac{213}{4}=28.25$ moles

mole of $(C{H}_4+C_2{H}_6+C_3{H}_8+N_2)$ taken ( in $100g$ of mixture )

$=\frac{Wt.}{Mol.Wt.}=\frac{100}{\frac{84\times 16+10\times 30+3\times 44+3\times 28}{100}}=\frac{10000}{1860}=5.367$

$\therefore 100$ moles of mixture produces $28.25$ moles

$\therefore 1$ ---------------------------------------- $\frac{28.25}{100}$

$\therefore 5.376$ ----------------------------------- $\frac{5.376\times 28.25}{100}$

$\therefore Wt.$ of $C_4{H}_6$ produced = mole $\times Mol.Wt.=1.518\times 54=82g$