Question

What mass of natural gas (CH4) must you burn to emit 263kJ of heat?

CH4(g) + 2O2(g)--> CO2(g) + 2H2O(g)
deltaHrxn= -802.3kJ

Answer 1

Treat the heat evolved as a stoichiometric product of the reaction you quoted.

Explanation:

Combustion of methane, which you have kindly provided:

`CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(g)` `;DeltaH = -802.3 kJ mol^-1`. Is it balanced? I think so.

The enthalpy change is quoted per mole of reaction as written. That is combustion of 1 mol of methane to give stoichiometric carbon dioxide and water vapour generates `-802.3` `kJ` precisely. So I could write the reaction this way, and treat the energy as a product (which indeed it is!).

`CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(g) + 802.3` `kJ`. So if `263` `kJ` are generated then `263/802.3` `(kJ)/(kJ)` `xx` `1*mol` methane were combusted. So this is something more than 1/4 a mole of methane combusted, approx. 5-6 g. Are we clear?