In a combustion reaction, 30g of propane burns in air to produce 70g of carbon dioxide.calculate the percent yield of carbon dioxide?

1 Answer
Apr 9, 2018

#77.8%#

Explanation:

balanced reaction:

#C_3H_8 + 5O_2 rarr 3CO_2 + 4H_2O#

the relative formula mass of #C_3H_8# is #(12 * 3) + (8 * 1)#, which is #44#.

the mass of propane burnt is #30g#.

number of moles = mass / molar mass

#= (30g) /( 44 g//mol)#

#= 0.682 mol# (3s.f.)

the number of moles of propane burnt is approximately #0.682#.

the reaction forms #3# molecules of carbon dioxide as a product, for every molecule of propane burnt.

this means that theoretically, #3 * 0.682# moles of #CO_2# would have been formed in the reaction.

the relative formula mass of #CO_2# is #(12 * 1) + (16 * 2)#, which is #44#.

using the same equation mass = number of moles * molar mass, the theoretical mass of #CO_2# formed would be

#(3 * 0.682)mol * 44 g//mol#, which is #90.02400..# grams.

we are given that the actual mass of #CO_2# formed is #70g#.

the percentage yield is calculated by dividing the actual mass of a product formed by the theoretical mass formed, and multiplying that ratio by #100%#.

here, the percentage yield is #70/90.02400 * 100%#,

which is #77.8%# (#3# s.f.)