Gas Stoichiometry Question

Question

Air is 21.0% by volume oxygen molecules, and C3H8 produces CO2 and H2O on combustion. What volume of air at 1.01 atm and 20.0 degrees Celsius would be required to burn 38.8 kg of propane (C3H8)?

Answer 1

Stoichiometry:

C3H8  +  5O2  -->  3 CO2  +  4 H2O

PV  =  nRT ,  V = nRT/P,

(38.8)*10^3 g  C3H8  =  879.9 mole C3H8 reacts with

                                5*(879.9) mole O2  = 4399.7 mole O2

                    V =  [4399.7 mole O2 * (.082 L atm/mole °K) * (293 °K)]/1.01 atm  =  104550 L O2.

        104660 L 02 * (1.00 L air/.210 L 02) =  498000 L of atmospheric air we breathe.

 

Answer 2

Now, let get started with an balanced combustion reaction: C₃H₈ + 5O₂ => 3CO₂ + 4H₂O

The pathway to do this is: mol of C₃H₈ -> mol of O₂ -> Volume of O₂ -> Volume of air.

Mol of C₃H₈ = Mass/Molar Mass = 38.8*10^3(g)/44(g/mol) = 881.82 mol

Mol of O₂ = 5*mol of C₃H₈ /1 = 5*881.82=4409 mol

Volume of O₂ = nRT/P = 4409*0.08206*(20+273)/1.01= 104960.7 L

Volume of air = volume of O₂ / 21% = 500,000 L

The final answer is 5.0*10^5 (L)

Answer 3

The first step in this problem is determining how much oxygen we need to burn 38.8 kg of propane.  Once we know how much oxygen we need in moles, we use the ideal gas law to convert moles of oxygen to a volume.  At that point, we can find the total volume of air.

Step one: How much oxygen do we need to burn 38.8 kg of propane?

  Combustion reactions use oxygen and produce CO2 and H20.  We need to write out the equation for the combustion of propane and determine its stoichiometry.

  C3H8 + 5O2 -> 3CO2 + 4H20 (since oxygen is in almost every molecule, balance all other atoms first)

  Now that we have the stoichiometric equation, we can determine the moles of oxygen we need.

  38.8 kg C3H8 * (1000g/1 kg) = 38800 g C3H8

  38800 g C3H8 * (1 mol C3H8/44.1 g C3H8) = 880 mol C3H8

  From the stoichiometric equation:

      880 mol C3H8 * (5 mol O2/1 mol C3H8) = 4400 mol O2

Step 2: What volume does 4400 mol O2 take at 1.01 atm and 20.0 degrees Celsius?

    Ideal gas law: PV=nRT

    P=1.01 atm; V=unkown;  n=4400 mol; R (gas constant)=0.082 L*atm/(K*mol); T=20*C+273=293 K

    1.01V=4400*0.082*293

    1.01V=105714.4

    V=104668 L of O2

Step 3:  Air is 21% by volume oxygen.

    Air*.21=104668 L of O2

   104668/.21=498418

Answer: It takes almost 500 kL of air to burn propane

Answer 4

Step 1: Write a balanced chemical equation.

C₃H₈ + 5 O₂ => 3 CO₂ + 4 H₂O

Step 2: Find out how much oxygen is needed to consume 38.8 kg of C₃H₈. We will only calculate this to moles because we'll be using the ideal gas equation later.

(38800 g C₃H₈) * (1 mol C₃H₈ / 44.1 g C₃H₈) * (5 mol O₂ / 1 mol C₃H₈) = X mol O₂. (4399 mol)

Step 3: Find out the volume of oxygen using V = (nRT)/P.

V_X = X moles O₂ * [.0821 (L*atm)/(mol*K)] * (293.15 K)/(1.01 atm).  (Roughly, 104700 L.)

Step 4: Divide the volume of oxygen by its percentage of air to find the total volume of air.

V_air = V_X / 0.21

Step 5: To reduce rounding errors, substitute all terms in and calculate once.

V_air = [(38800 / 44.1 * 5) * (.0821 * 293.15 / 1.01)] / 0.21

        = 499178.1 L air

Step 6: Mind the significant figures. We started with three sigfigs, so we must end with three sigfigs.

499 kL air