The reaction, `2C+O_(2)rarr2CO` is carried out by taking `24 g` of carbon and `96 gO_(2)`, find out:
(a) which reactant is left in excess?
(b) How much of it is left?
(c ) How many mole of `CO` are formed?
(d) How many `g` of other reactant should be taken so that nothing is left at the end of reaction?

Explanation

`underset(=24g)underset("2mol")(2Mg)2C(s)+underset(32g)underset(1"mol")(O_(2)) to underset(56g)underset(2"mol")(2CO(g))`
Let carbon be completely consumed.
24g carbon give 56g CO.
Let `O_(2)` give 56g of CO.
`therefore 96"g"O_(2) "will give" (56)/(32)xx96g CO=168"g CO"`
Since, carbon gives least amound of product, i.e. `56g of CO or 2` of mole CO, hence carbon will be the limiting reactant.
Excess reactant is `O_(2)`
Amound of `O_(2)` used=56-24=32g
Amount of `O_(2)` left=96-32=64g
32g of `O_(2)` react with 24g carbon
`therefore 9g of O_(2) ` will react with 72g carbon
Thus, carbon should be taken 72g so that nothing is left at the end of the reaction.