If 538 mol of octane combusts, what volume of carbon dioxide is produced at 14.0°C and .995 atm?

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Answer 1 · 2016-05-08T08:47:01

Over one hundred thousands litres of carbon dioxide gas are evolved.

Balanced combustion equation:

$C_8H_18(l) + 17/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)$

So, given that $538*mol$ of octane combust, clearly, there are $8xx538*mol$ $CO_2$ evolved under complete combustion.

$V=(nRT)/P$

$=$

$(8xx538*cancel(mol)xx0.0821*L*cancel(atm)*cancel(K^-1)*cancel(mol^-1)xx287*cancelK)/(0.995*cancel(atm))$

$??*L$

What is the volume in $m^3$ ?