Question

At standard temperature (298 K) and pressure (1 atm), what volume of air would be required to burn 100 g of propane? Assume that air consists of 21% oxygen.

please give a solution

Answer 1

Answer –

We are given, mass of propane, C₃H₈ = 100 g , 21 % O₂ in the air

Step 1) Write the balanced equation for the combustion of propane -

C₃H₈ + 5O₂ ---> 3CO₂ + 4H₂O

Step 2) Calculate the moles of propane

Moles of C₃H₈ = 100 g / 44.0462 g.mol^-1

                         = 2.27 moles

Step 3) Calculate the moles of oxygen

From the balanced equation –

1 moles of C₃H₈ = 5 moles of O₂

So, 2.27 moles of C₃H₈ = ?

= 2.27 moles of C₃H₈ * 5 moles of O₂ / 1 moles of C₃H₈

= 11.35 moles of O₂

Step 4) Calculate the volume of the oxygen gas

We are given P and T and we calculated the moles of oxygen

So, PV= nRT

V = nRT/P

     = 11.35 moles * 0.0821atm.L.mol^-1.K^-1 * 298 K / 1 atm

      = 277.7 L

Step 5) Calculating the volume of air

We are given that 21 % of O₂ in air

So, 100 % air = 21 % O₂

Means we assume 21 L of O₂ = 100 L of air

So, 277.7 L O₂ = ?

= 1322.5 L of air

So volume of air would be required to burn 100 g of propane is 1322.5 L