As from rection $A+2 B \rightarrow C$
It is clear that $1$ mole of $A$ reacts with $2$ moles of $B$ to form $1$ mole of $C$.
$\therefore 4$ moles of $A$ will react with $2 \times 4=8$ moles of $B$ to form $4$ moles of $C$
Here, $B$ is limiting reagent.
$1$ mole of $A$ remains unreacted.