If a sample of butene (c4h8) that has a mass of 259.9 g is combusted in excess oxygen, what is the mass of co2 that is produced?

c4h8 + 6 o2 ⟶ 4 co2 + 4 h2o

The balanced chemical reaction is expressed as:

C4H8 + 6 O2 ⟶ 4 CO2 + 4 H2O

To determine the mass of carbon dioxide that is produced from the reaction, we use the initial amount of the limiting reactant which is butene in this case. Also, from the relation of the substances in the reaction we can relate CO2 and the amount of the reactant. We calculate as follows:

moles C4H8 = 259.9 g C4H8 ( 1 mol C4H8 / 56.12 g C4H8 ) = 4.63 mol C4H8
moles CO2 = 4.63 mol C4H8 ( 4 mol CO2 / 1 mol C4H8 ) = 18.52 mol CO2
mass CO2 = 18.52 mol CO2 ( 44.01 g / 1 mol ) = 815.27 g CO2 produced

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The answer is c - the principal of original lateral continuity.