Question

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

Answer 1

`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`