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

Question

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

Ellie Andrews · Accepted Answer

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

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#

Jose Ayala · Answer

undefined To find the mass of propane required to yield 76000 kJ of energy, use the formula:

$$ \text{Mass} = \frac{\text{Energy}}{\Delta H_\text{combustion}} $$

Substitute the given values:

$$ \text{Mass} = \frac{76000 \text{ kJ}}{2220 \text{ kJ/mol}} $$

$$ \text{Mass} = 34.23 \text{ mol} $$

Since the molar mass of propane ($C_3H_8$) is approximately 44.1 g/mol:

$$ \text{Mass} = 34.23 \text{ mol} \times 44.1 \text{ g/mol} = 1510.7 \text{ g} $$