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(2xx12=24g)(2C)+underset(32g)(O_(2))rarrunderset(2xx28=56g)(2CO)`
24 g of C will require oxygen `(O_(2)) = 32 g`
But `O_(2)` actually available = 96 g
(a) Therefore, reactant left in excess `= O_(2)`
(b) Mass of `O_(2)` left unreacted `= 96 - 32 = 64 g`
(c) For reacting with 32 g of `O_(2)`, carbon needed = 24 g
For reacting with 96 g of `O_(2)`, carbon needed `= ((24g))/((32g))xx(96g)=72g`.