Calculate the amount of aluminium, required to get 1120 kg of iron from the given chemical equation.
`2Al_((s))+Fe_(2)O_(3(s))rarrAl_(2)O_(3(s))+Fe_((s))" "` (atomic mass of Al = 27U, Fe = 58U and O = 16 U)

Explanation

`{:(2Al_(s),+,Fe_(2)O_(3(s)),rarr,Al_(2)O_(3(s)),+,Fe_((s))),((3xx27)U,+,(2xx56+3xx16)U,rarr,(2xx27+3xx16)U,+,(2xx56)U),(54U,+,160U,rarr,120U,+,112U),("2moles",+,"1mole",rarr,"1mole",+,"2moles"),("54 gm",+,"160gm",rarr,"120 gm",+,"112 gm"):}`
So, aluminium `rarr` iron
54gm `rarr` 112 gm
x gm? `rarr(1120xx1000)gm`
`therefore" "x=((1120xx1000)gmxx54gm)/(112gm)=10000xx54gm=540000gm=540kg`
`therefore" "`To get 1120 kg of iron we have to use 540 kg of aluminium.