In the combustion of methane, how many grams of methane are needed to produce 150 L of carbon dioxide?

The balanced equation for this reaction is:

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

Given Amount

The given amount is: 150 L `~CO_2` .``

Conversion Factors

We will need to use the following mole conversion factors to solve this problem.

1 mole = 22.4 L `~CO_2`

1 mole = *molar mass = 16.042 g `~CH_4`  

*The molar mass is calculated by multiplying the atomic mass of each element in the compound by its subscript and adding the resulting products together.

Mole Ratio

We will also need the mole ratio between `~CO_2` and `~CH_4` . A mole ratio is the ratio between the coefficients of two substances in a chemical reaction.

The coefficient of `~CO_2` is 1.

The coefficient of `~CH_4` is 1.

Therefore, the mole ratio is: 1 mole `~CO_2` = 1 mole `~CH_4`

Stoichiometry Calculation

The calculation will take the general form:

   given amount x conversion factor x mole ratio x conversion factor

Therefore, 

   150 L `~CO_2` x (1 mole/22.4 L) x (1 mol `~CH_4` /1 mol `~CO_2` ) x (16.042 g/1 mole)

      = 107 g `~CH_4`

Notice that the conversion factors and mole ratio are oriented such that all units cancel out except for the final unit of grams.