Question

In the equation `2C_2H_6 + 7O_2 -> 4CO_2 + 6H_2O`, if 15 g of `C_2H_6` react with 45 g of `O_2`, how many grams of water will be produced?

Answer 1

`"Moles of ethane"` `=` `(15*g)/(30.07*g*mol^-1)=0.499*mol`.

I gets `27*g` water product............

Explanation:

`"Moles of dioxygen"` `=` `(45*g)/(32.00*g*mol^-1)=1.41*mol`.

Clearly, there is insufficient dioxygen for complete combustion; i.e. complete combustion requires `1.75*mol` `O_2`. We ASSUME incomplete combustion, i.e.

`"COMPLETE COMBUSTION GIVES...."`

`C_2H_6(g) + 7/2O_2(g) rarr 2CO_2(g) + 3H_2O(l)`

`"INCOMPLETE COMBUSTION GIVES...."`

`C_2H_6(g)+3O_2(g)rarrCO_2(g) + CO(g)+3H_2O(l)`

`"OR.........."`

`C_2H_6(g)+2O_2(g)rarrC(s)+ CO(g)+3H_2O(l)`

(What would we obtain? This would be the subject of experiment. Certainly incomplete combustion occurs in the internal combustion engine, and in diesels.) In each case, however, combustion of ethane, complete, or incomplete, gives rise to...........

`"THREE EQUIV of WATER PER EQUIV ETHANE."`

And thus if there are `0.5*mol` ethane reactant, there will be `1.5*mol` water product.........a mass of `27*g`.

Good question, which I am stealing for my A2 class.