The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals. (Atomic mass of Al = 27 u Atomic mass of 0 = 16 u) `2Al + Fe_(2)O_(3) to Al_(2)O_(3)+ 2Fe`, If, in this process, 324 g of aluminium is allowed to react with 1.12 kg of ferric oxide.
(i) Calculate the mass of `Al_(2)O_(3)` formed.
(ii) How much of the excess reagent is left at the end of the reaction?

Explanation

`underset(54g)(2Al) + underset(160g)(Fe_(2)O_(3)) to underset(102g)(Al_(2)O_(3)) + underset(112g)(2Fe)`
As per balanced equation 54 g Al is required for 112 g of iron and 102 g of `Al_(2)O_(3)` .
54 g of Al gives 102 g of `Al_(2)O_(3)`
`:.` 324 g of Al will give `102/54 xx324= 612` g of `Al_(2)O_(3)` .
(ii) 54 g of Al requires 160 g of `Fe_(2)O_(3)` for welding reaction.
324 g of Al will require`160/54xx324` = 960 g of `Fe_(2)O_(3)` .
`:.` Excess `Fe_(2)O_(3)` - Unreacted `Fe_(2)O_(3)` = 1120 - 960 = 160 g
160 g of excess reagent is left at the end of the reaction.