Step-by-step explanation:
Liquid octane will burn according to the following balanced equation:
2 CH₃(CH₂)₆CH₃ (l) + 25 O₂ (g) ----> 16 CO₂ (g) + 18 H₂O (g)
We have to find the mass of CO₂ that is produced when 88.0 g of octane are reacted with 277. g of oxygen gas. We will have to find the limiting reagent. The first step to do that is to convert the masses into moles using the molar masses of the reactants.
molar mass of CH₃(CH₂)₆CH₃ = 114.24 g/mol
molar mass of O₂ = 32.0 g/mol
moles of CH₃(CH₂)₆CH₃ = 88.0 g * 1 mol/(114.24 g)
moles of CH₃(CH₂)₆CH₃ = 0.770 moles
moles of O₂ = 277. g * 1 mol/(32.0 g)
moles of O₂ = 8.66 moles
So we are mixing 0.770 moles of octane with 8.66 moles of oxygen gas. Now we have to find the moles of carbon dioxide that can be theoretically produced by each reactant as they were reacting with excess of the other one. Let's begin with octane.
2 CH₃(CH₂)₆CH₃ (l) + 25 O₂ (g) ----> 16 CO₂ (g) + 18 H₂O (g)
If we pay attention to the coefficients of the equation we will see that 2 moles of octane will produce 16 moles of carbon dioxide (when reacting with oxygen excess). So the molar ratio between them is 2 to 16. We can use that relationship to find the number of moles of carbon dioxide that can be produced from 0.770 moles of octane.
2 moles of CH₃(CH₂)₆CH₃ : 16 moles of CO₂ molar ratio
moles of CO₂ = 0.770 moles of CH₃(CH₂)₆CH₃ * 16 moles of CO₂/( 2 moles
moles of CH₃(CH₂)₆CH₃)
moles of CO₂ = 6.16 moles
We can do something similar with the moles of O₂.
25 moles of O₂ : 16 moles of CO₂ molar ratio
moles of CO₂ = 8.66 moles of O₂ * 16 moles of CO₂/( 25 moles
moles of O₂)
moles of CO₂ = 5.54 moles ----> oxygen is the limiting reagent
We found that 0.770 moles of octane can produce 6.16 moles of carbon dioxide and 8.66 moles of oxygen gas can produce 5.54 moles of carbon dioxide. So, we can say that oxygen is the limiting reagent since it is limiting the reaction.
Finally we can convert the moles of carbon dioxide into moles using the molar mass of it.
¨
molar mass of CO₂ = 44.01 g/mol
mass of CO₂ = 5.54 moles * 44.01 g/mol
mass of CO₂ = 243.81 g
mass of CO₂ = 244 g
Answer: the theoretical yield of carbon dioxide formed is 244 g.