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?

The reaction, `2C+O_(2)rarr2CO` is carried out by taking `24 g` of car

1.	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?
Answer» `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.

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