Answer:
98.14 g of CO2
Explanation:
So to solve this problem we are going to use the ideal gas law or PV=nRT.
So we are actually only missing one value here in this question if we pay attention to the wording!
We are given a volume of 50L so we have that
At STP pressure and temperature are constant
STP means:
A temperature of 273.15K / 0 degrees Celsius as well as 1 atm of pressure.
So that means
T = 273.15K
P = 1 atm
R is the ideal gas constant which has various values but in this case will have a value of 0.08206 for all of our units to cancel correctly.
So lets take PV=nRT and isolate for n (the number of moles)
We want the number of moles so we can find the number of moles of methane that were burned and then we can convert this molar amount to a molar amount of CO2 produced.
So we have:
[tex]\frac{PV}{RT}= n[/tex]
Lets plug what we know in!
[tex]\frac{(1atm)(50L)}{(0.08206)(273.15K)} = n[/tex]
This means n = 2.23 mol of Methane
Now lets look at our chemical equation. We can see that methane and CO2 have the same coefficient so we dont need to do anything and we can say that there is therefore 2.23 mol of CO2 produced.
Now all thats left is to convert this amount of moles to grams using our molar mass of CO2. The molar mass of CO2 is 44.01g.
Heres the conversion:
[tex]\frac{2.23mol(CO2)}{1} (\frac{44.01g}{mol(CO2)} )[/tex]
This equals 98.14 g of CO2. Hence the amount of CO2 produced is 98.14g.
Let me know if you have any questions regarding this!
