mixing 108 g of propane, C3H8, with 232 L of O2 at STP. How many grams of CO2 will be produced? The balanced equation for this reaction is: C3H8 + 5 O2 →3 CO2 + 4 H2O

Question

I am taking an online chemistry class and the teacher is not much help with anything... I was hoping someone here could explain it to me?

Answer 1

I'll try to explain this best I can.

First, always write a correctly balanced equation:

C₃H₈ + 5 O₂ ==> 3 CO₂ + 4 H₂O

Next, find the limiting reactant (C₃H₈ or O₂):

moles C₃H₈ present = 108 g x 1 mol/44.1 g = 2.45 moles

moles O₂ present = 232 L x 1 mol/22.4 L = 10.4 moles (recall that at STP 1 mole of any ideal gas = 22.4 L)

One easy way to find limiting reactant is to divide the moles of each by their corresponding coefficient in the balanced reaction. Here you have 2.45 moles C₃H₈/1 = 2.45. For O₂ you have 10.4/5 = 2.08 so O₂ would be limiting. Another way is to calculate moles of product formed from each reactant.

Now that we have the limiting reactant, we can find the moles of CO₂ produced, and then the mass of CO₂:

10.4 moles O₂ x 3 moles CO₂/5 moles O₂ = 6.24 moles CO₂ produced

mass CO₂ = 6.24 moles CO₂ x 44 g/mole = 275 g CO₂

Answer 2

Chloe,

Here are the steps:

First, find how many moles of C3H8 and O2 you have to work with.

Second, use one of them to figure the limiting reactant.

Third, use the limiting reactant to convert moles of that substance to moles of CO2.

Lastly, convert moles of CO2 to grams CO2.

(108g C3H8)(1 mol/44.11g) = 2.45 mol C3H8

Use PV=nRT to get your moles of O2

Rearrange to PV/RT

(1atm)(232L)/.0821l-atm/mol-K(273K) = 10.4 mol O2

It's hard to see the equation when I can't really draw the fractions, but the atm and L of O2 are on the top and R and K are on the bottom.

I started with C3H8 but you could have started with the O2:

(2.45 mol C3H8) x (5 mol O2/1 mol C3H8) =12.3 mol O2. You only have 10.4, so O2 is the limiting reactant.

(10.4 mol O2) x (3 mol CO2/5 mol O2) = 6.24 mol CO2

(6.24 mol CO2)(44.03 g CO2/mol) = 275 grams CO2. That's rounded for sig fig.

Hope that helps! You can message me by looking for my name in Wyzant or by going to www.wyzant.com/tutors/acatutoring

Barb