Propane (C3H8) burns in excess oxygen according to the following reaction: C3H8(

Question

Propane (C3H8) burns in excess oxygen according to the following reaction: C3H8(g) + 5 O2(g) -> 3 CO2(g) + 4 H2O(l)
What amount of propane will produce 250g of CO2 if the percent yield is 93.1%?
_______g? Propane (C3H8) burns in excess oxygen according to the following reaction: C3H8(g) + 5 O2(g) -> 3 CO2(g) + 4 H2O(l)
What amount of propane will produce 250g of CO2 if the percent yield is 93.1%?
_______g? C3H8(g) + 5 O2(g) -> 3 CO2(g) + 4 H2O(l)
What amount of propane will produce 250g of CO2 if the percent yield is 93.1%?
_______g?

Explanation / Answer

1mol C3H8 produces 3 mol CO2
Molar mass CO2 = 44g/mol
moles in 250g CO2 = 250/44 = 5.682 mol CO2
Because yield is 93.1% , you must plan to make: 5.682*100*93.1 = 6.103 mol CO2

You will require 6.103/3 = 2.034 mol C3H8
Molar mass C3H8 = 44g/mol
Mass of 2.034 mol C3H8 = 2.034*44 = 89.5g C3H8 required.

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