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Limiting reagent problem!  If 1.00L of methane is reacted with 1.00L of oxygen...

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fantabulous987 | Student, Grade 11 | eNoter

Posted May 28, 2013 at 3:33 AM via web

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Limiting reagent problem! 

If 1.00L of methane is reacted with 1.00L of oxygen according to the following equation...

CH4(g) + 2O2(g) => CO2(g) + 2H2O(g)

Determine the volume of carbon dioxide produced under the same conditions of temperature and pressure. 

I don't understand how we can find the limiting reagent without having the moles of the product. and we can't convert the volume to moles because we don't have enough information for the ideal gas law (PV=nRT).

if someone could walk me through how to solve this problrm please! :)


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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted May 28, 2013 at 5:18 AM (Answer #1)

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It will be quite confusing in dealing with this type of problem. However, we can see that the problem says "same conditions of temperature and pressure" and since the volume of the two reactants are the same, the number of moles of CH4 and O2 will be the same. 

Now we know that the moles of CH4 and O2 are the same, we should determine the amount of the products that is produced in the reaction. 

Looking at the balanced chemical reaction,

`CH_(4)_(g) + 2O_(2)_(g) -> CO_(2) _(g) + 2H_(2)O_((g))`

we can see that two liters of O2 is needed to react with the 1 liter of CH4. Since both of them have amount, the first that will be consumed is the 1L O2. Therefore the limiting reactant is the O2. Only half of the CH4 will be used. Meaning, 0.5 L of excess CH4 will remain.

`1 L O_(2) * (1 L CO_(2))/(2 L O_(2)) = 0.5 L CO_(2)`

`1 L O_(2) * (2 L H_(2)O)/(2 L O_(2)) = 1 L H_(2)O`


Total volume = volume of excess + volume CO2 + volume H2O

Total volume = 0.5 + 0.5 + 1 = 2.0 L under the same conditions.


Another way to solve this is to assign a specific value for temperature and pressure. For example, use the value at STP or standard temperature and pressure (T = 273.15 K; 1 atm). You will also arrive at the same answer. 


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