Question

Question #e8ef9

Answer 1

`CH_4 + 2O_2 -> CO_2 + 2H_2O`

Explanation:

24.055 L of methane gas (density of methane 0.668 g/L at normal temperature and pressure). Mass of methane `M=0.668times24.055 = 16.08` grams. Methane molecular weight is 16 grams. It means you have only 1.004 mole of methane. Your products are carbondioxide (1.004 mole) and water vapor (1.004 mole). Therefore, you will have 44.19 grams of `CO_2` after burning 1.004 mole of methane.

Volume of carbondoxide can be computed by:

`V=(ntimesRtimesT)/P = (1.004times0.08206times298.15)/1 = 24.56` Liters.

Density of gases:
http://www.engineeringtoolbox.com/gas-density-d_158.html