CH4 + O2 ===> CO2 + H2O: How many grams of carbon dioxide will be produced when 8.0g of methane (CH4) react with 15.9 g of oxygen?

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jeew-m eNotes educator | Certified Educator

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Molar mass

`CH_4 ` = 16g/mol

`O_2 ` = 32g/mol


`CH_4+2O_2 rarr CO_2+2H_2O`


Number of `CH_4` moles added `= 8/16 = 0.5 mol`

Number of `O_2` moles added `= 15.9/32 = 0.497 mol`


Mole ratio

`CH_4:O_2 = 1:2`


Since the amount of `O_2` moles needed for reaction is twice the `CH_4` moles all `CH_4` cannot react and form `CO_2` .

Amount of `CH_4` reacted `= 0.497/2 = 0.248`


Mole ratio

`CH_4:CO_2 = 1:1`


So the amount of `CO_2` produced is 0.248 moles





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ncchemist eNotes educator | Certified Educator

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You are only allowed to ask one question at a time so I edited down accordingly. First we must balance the chemical equation you have given for the combustion of methane:

CH4 + O2 --> CO2 + H2O

There are 4 hydrogens on the left so there should be 4 hydrogens on the right. That means we need a 2 in front of the water. As a result, we need to add a 2 in front of the O2 to balance out the oxygens. That gives the balanced equation as:

CH4 + 2O2 --> CO2 + 2H2O

Now let's convert the grams of methane and oxygen to moles and see which one is the limiting reagent.

8 g methane (1 mole/16 g) = 0.5 mole methane

15.9 g O2 (1 mole/32 g) = 0.5 mole O2

But from the equation we know that we need 2 moles of oxygen for every mole of methane so that makes oxygen the limiting reagent.  So 0.5 moles of O2 will form 0.25 moles of CO2 (2:1 ratio of O2:CO2).  So now convert moles of CO2 to grams:

0.25 moles CO2 (44 g/1 mole) = 11 grams CO2

So 11 grams of carbon dioxide are produced.

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