| Why can it be assumed that that the heat of the reaction is the same thing as the heat as one part of the components of the reaction
C3H8(g)+5O2(g) yields 3CO2(g)+4H20
delta H is -2044KJ
what mass of C3H3 is necessary to heat 1.8L of water from 25 to 100 degree Celsius ? assume only 15% of heat emitted from combustion reaction to heat up the water

I understand that (1800ml)(4.18J/g Celsius)(75) to find heat of water
I don't understand what to do next

Question

| Why can it be assumed that that the heat of the reaction is the same thing as the heat as one part of the components of the reaction
C3H8(g)+5O2(g) yields 3CO2(g)+4H20
delta H is -2044KJ
what mass of C3H3 is necessary to heat 1.8L of water from 25 to 100 degree Celsius ? assume only 15% of heat emitted from combustion reaction to heat up the water

I understand that (1800ml)(4.18J/g Celsius)(75) to find heat of water
I don't understand what to do next

anonymous · Answer

Ahhh, well first off, $C_3H_8$ is propane gas. Oxygen is in the air all around us, so we don't have to provide it to cause this reaction to work since it's in excess, so that's why they just say "Combustion of $C_3H_8$" because at the end of the day that's what's burning and is the important part of the reaction. So hopefully that answers your first question about why they are only talking about one part of the reaction.

Next up, it looks like you're on the right path, the only problem is I don't know if you've been given the specific or molar enthalpy. The reason being that they have two slightly different meanings so if you see any other symbols or letters around where they have $\Delta H$ I can help you figure this out.