If $6$g of $Fe$ reacts with $1$g of oxygen, the amount of $FeO$ formed will be $(Fe=56)$, if only FeO is formed as a product.

The correct option is A $4.5$g
$6g$ of $Fe$ contains=$\frac{6}{56}$ moles of $Fe$

$1g$ of $O_2$ contains= $\frac{1}{32}$ moles of $O_2$

moles $Fe+1/2O_2⟶FeO$

at $t=0$ $\frac{6}{56}$ $\frac{1}{32}$

$\frac{1}{2}$ mole of $O_2$ is reacting with $1$ mole of $Fe$. Thus $\frac{1}{32}$ mole of $O_2$ reacts with $2\times \frac{1}{32}=\frac{1}{16}$ mole of $Fe$. Thus $O_2$ is limiting reagent here, as$\frac{1}{2}$ mole of $O_2$ gives $1$ mole of $FeO$. Thus $\frac{1}{32}$ mole of $O_2$ will give $\frac{2}{32}$ mole of $FeO$

Thus amount of $FeO$ in $\frac{1}{16}$ mole produced is $\frac{1}{16}\times 72=4.5g$

Thus $4.5g$ of $FeO$ will be produced.