The combustion of propane...?

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1 Answer
Jul 5, 2018

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 ...