Question

The combustion of propane...?

Answer 1

You have the stoichiometry...

Explanation:

`C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l) + Delta`

`"Moles of propane"=(2200*g)/(44.1*g*mol^-1)=49.9*mol`

And clearly, by the stoichiometry if `50*mol` propane are COMPLETELY combusted...`150*mol` carbon dioxide results, and `250*mol` dioxygen are required...

i.e. a mass of `49.9*molxx5xx32.00*g*mol^-1~=8*kg` ...