2 Answers | Add Yours
`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`
`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`
`CH_4:CO_2 = 1:1`
So the amount of `CO_2` produced is 0.248 moles
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.
We’ve answered 302,420 questions. We can answer yours, too.Ask a question