# 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

Nov 14, 2015

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

#### Explanation:

Combustion of methane, which you have kindly provided:

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right)$ ;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 $- 802.3$ $k J$ precisely. So I could write the reaction this way, and treat the energy as a product (which indeed it is!).

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right) + 802.3$ $k J$. So if $263$ $k J$ are generated then $\frac{263}{802.3}$ $\frac{k J}{k J}$ $\times$ $1 \cdot m o l$ methane were combusted. So this is something more than 1/4 a mole of methane combusted, approx. 5-6 g. Are we clear?