#DeltaH_"combustion"^@# #"propane"=2220*kJ*mol^-1#. What mass of propane is required to yield #76000*kJ# of energy?

1 Answer
anor277
Mar 14, 2017

#DeltaH""_"rxn"^@# is ALWAYS written per mole of reaction as written. We need approx. #1.5*kg# of propane.........

Explanation:

And thus the complete combustion of 1 mole of propane, i.e. #44.1*g#, yields #2200*kJ# as shown:

#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(g) +Delta#

#"OR.............."#

#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(g) +2200*kJ#

We put the energy output on the PRODUCT side inasmuch the combustion equation is EXOTHERMIC (the negative sign of #DeltaH# assures us that this is the case). And thus, in effect, we use #Delta# as a virtual product in the reaction, and we solve the following quotient:

#(76000*kJ)/(2200*kJ*mol^-1)=34.6*mol#, by which we mean #34.6*mol# of reaction as written.

And thus we require #34.6*mol# propane, i.e. #34.6*molxx44.1*g*mol^-1=1523.5*g#

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