Question

If 3.14 moles of oxygen reacts with butane, how many grams of butane is needed?

Answer 1

Of course the butane COULD combust incompletely...

Explanation:

...to give `C(s)` and `CO(g)` as the products of incomplete combustion. But here we must ASSUME complete combustion in order to address the question:

`C_4H_10(g) + 13/2O_2(g) rarr 4CO_2(g) + 5H_2O(l)`

And so if `3.14*mol` dioxygen react...then we get with respect to butane...

`3.14*molxx2/13xx58.12*g*mol^-1=28.1*g`.

Most of the time, the shorter alkanes (i.e. butane and lower) combust cleanly...the longer chains, oils and diesels tend to combust quite incompletely.