given this reaction: 2CO(g) + O2(g)---2CO2(g) what volume of CO2 will be produced when 8.45 liter of CO reacts with enough oxygen? I thought the answer would be 8.45 liters of CO2 because there was a ratio in the balanced equation of 2CO yielding 2CO2, so I thought the ration would be 2 to 2 or 1 to 1, but according to the answer book, that is wrong. I don't understand why
I think your answer is correct and the answer in the book is wrong, unless I'm missing something here.
To determine the volume of CO2 produced when 8.45 liters of CO reacts with enough oxygen, we need to use the stoichiometry of the balanced chemical equation.
The balanced chemical equation provided is:
2CO(g) + O2(g) → 2CO2(g)
From the balanced equation, we can interpret that 2 moles of CO react to produce 2 moles of CO2. However, in this case, we have the volume of CO instead of moles.
To convert the volume of CO to moles, we need to use the ideal gas law. The ideal gas law equation is given by:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Assuming the reaction occurs at the same temperature and pressure, we can rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Now, let's calculate the number of moles of CO using the given volume:
n(CO) = (PV) / RT
Substituting the given values:
P = pressure (assume constant)
V(CO) = 8.45 L
R = ideal gas constant
T = temperature (assume constant)
Once we know the number of moles of CO, we can use the stoichiometry of the balanced chemical equation to determine the number of moles of CO2 produced. From the balanced equation, we know that 2 moles of CO react to produce 2 moles of CO2.
Therefore, the number of moles of CO2 produced will also be equal to the number of moles of CO present.
Finally, we can convert the moles of CO2 to volume using the ideal gas law again:
V(CO2) = n(CO2) * RT / P
Substituting the calculated value for n(CO2), the ideal gas constant (R), temperature, and pressure, we can find the volume of CO2 produced.