Friedel–Crafts alkylation

Ethylbenzene, $C_6H_5CH_2CH_3$, needs an ethyl group to be attached to the benzene ring.
Ethylene in the presence of $AlCl_3$ (with a trace of HCl) gives the ethyl carbocation, $CH_3CH_2^+$, which attacks benzene (electrophilic substitution):
$CH_2=CH_2 + C_6H_6 \xrightarrow{anhyd.\ AlCl_3,\ HCl} C_6H_5CH_2CH_3$
This is the industrial method for making ethylbenzene (for styrene manufacture).
Propene would give the more stable isopropyl carbocation and so form isopropylbenzene (cumene), not ethylbenzene.
Ethane is a saturated alkane and does not produce a carbocation with $AlCl_3$. Ethanol reacts with and deactivates $AlCl_3$, so it is not used with this catalyst.
So ethylbenzene is produced from $H_2C=CH_2 + C_6H_6$.