The combustion reaction of methane, Ch4, produces carbon dioxide, water and heat. If you burn 40.0 grams of methane in a counter that contains 128.0 grams of oxygen, how much carbon dioxide is...

The combustion reaction of methane, Ch4, produces carbon dioxide, water and heat. If you burn 40.0 grams of methane in a counter that contains 128.0 grams of oxygen, how much carbon dioxide is produced?

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jerichorayel | College Teacher | (Level 2) Senior Educator

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First, let us write the balanced chemical equation:

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

First we have to determine the limiting reagent to be able to know the amount of products that will be produced.

For `CH_4` :

`40.0 grams CH_4 * (1 mol e CH_4)/(16.04 grams) * (1 mol e CO_2)/(1 mol e CH_4)`

= 2.494 moles `CO_2`

 

For `O_2` :

`128 grams O_2 * (1 mol e O_2)/(32 grams) * (1 mol e CO_2)/(2 mol es O_2)`

= 2.00 moles `CO_2`

 

Looking at the results, the limiting reagent is the oxygen gas and therefore we will use the amount of `CO_2` derived from `O_2` .

`2.00 mol es CO_2 * (44.01 grams CO_2)/(1 mol e CO_2)`

= 88.0 grams `CO_2` -> answer

Sources:

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