1.00l of octane (M=114.23 g/mol, d=0.703 g/cm3) undergoes a combustion reaction with 5.0 L of oxygen at 25 celcius and 1.0 atm pressure. what mass of co2 (M=44.01 g/mol) is produced?

Answer:

               2.873 g of CO₂

Explanation:

This problem will be solved in two steps.

Step 1: Calculating mass of Octane:

Data Given:

                  Volume = 1 L = 1000 cm³

                  Density  =  0.703 g/cm³

                  Mass  =  ??

Formula Used:

                       Density = Mass ÷ Volume

Solving for Mass,

                       Mass  =  Density × Volume

                       Mass  =  0.703 g/cm³ × 1000 cm³

Putting Values,

                       Mass  =  703 g

Step 2: Calculating Mass of Oxygen:

Data:

                  Volume =  V  = 5.0 L

                  Temperature = T = 25 °C = 298.15 K

                  Pressure = P = 1.0 atm

                  Moles = n = ?

Assuming that the gas is acting as Ideal gas so, we will use Ideal gas equation i.e.

                 P V = n R T

Solving for n,

                  n = P V / RT

Putting values,

                  n = 1.0 atm × 5.0 L / 0.0821 atm.L.mol⁻¹.K⁻¹ × 298.15 K

                  n = 0.204 moles

As,

            Moles = Mass / M.Mass

So,

            Mass = Moles × M.Mass

            Mass = 0.204 mol × 16 g/mol       ∴ M.Mass of O₂ = 16g.mol⁻¹

            Mass = 3.26 g

Step 3: Calculating mass of CO₂:

The balance chemical equation is follow,

                            2 C₈H₁₈ + 25 O2 = 16 CO₂ + 18 H₂O

According to equation

        228.45 g (2 mol) of C₈H₁₈  reacts with = 799.97 g (16 mol) of O₂

So,

                       703 g of C₈H₁₈ will react with = X g of O₂

Solving for X,

                  X  =  703 g × 799.97 g ÷ 228.45

                  X  =  2461 g of O₂

While, we are only provided with 3.26 g of O₂. This means O₂ is the limiting reactant and will control the yield of the final product. Therefore,

According to balance equation,

          799.97 g (16 mol) of O₂ produced  =  704.152 g (16 mol) of CO₂

So,

   3.26 g (0.204 mol) of O₂ will produce  =  X g of CO₂

Solving for X,

                 X =  3.26 g × 704.152 g ÷ 799.97 g

                  X =  2.873 g of CO₂