Given the combustion of butene, C$4$H$8($g$)+6$O$2($g$)$rightwards arrow$4$CO$2($g$)+4$ H$2$O $($g$),$ determine the volume, to the nearest cubic metre, of CO$2$ produced, at STP$,$ when $99$ kg of butene are combusted. Take the molar mass of butene to be $56$ kg$/$kmol$.$ Include the units m$3$ in your answer.

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First lets start by clarifying the provided equation for the combustion of butene C4H8g 6O2g 4CO2g 4... View the full answer