For the reaction:

$F{e}_2{O}_3(s)+2Al(s)2Fe(l)+A{l}_2{O}_3(s)$

What mass of iron$($III$)$ oxide must be used to produce $25\mathrm{.}0g$ of iron?

$Fe=56\frac{g}{m}ol,=16\frac{g}{m}ol,$

$35\mathrm{.}7g$

$67\mathrm{.}2g$

$89\mathrm{.}5g$

$110g$