Question

How do you determine the mass of carbon dioxide produced when 0.85 g of butane reacts with oxygen according to the following equation: `2C_4H_10 + 13O_2 -> 8CO_2 10H_2O`?

Answer 1

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

`5.58xx10^-2 "moles of "CO_2" are produced".`

Explanation:

`"Moles of butane"` `=` `(0.85*g)/(58.12*g*mol^-1)` `=` `1.46xx10^-2mol`.

Given the stoichiometry, each moles of butane gives `4*mol` `CO_2` upon complete combustion.

And thus `"moles of "CO_2` `=` `4xx1.46xx10^-2mol=5.58xx10^-2mol`.

This is an equivalent mass of `5.58xx10^-2molxx44.01*g*mol^-1~=2.5*g`.

In my stoichiometric equation, I used an half equiv of dioxygen in order to balance the equation. I do this because it makes the 'rithmetic a bit easier.