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...

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jerichorayel's profile pic

Posted on

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


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