Answer to Question #136987 in General Chemistry for mica

Question

The burner in a gas grill mixes 24.0 volumes of air for every one volume of propane (C3H8) fuel. Given: like all gases, the volume that propane occupies is directly proportional to the number of moles of it at a given temperature and pressure. Air is 21.0% (by volume) O2.
II) How many moles of propane are needed to completely use up the oxygen in 24.0 moles of air?

I)Is the flame produced by the burner: fuel-rich (excess propane in the reaction mixture), fuel-lean (not enough propane), or stoichiometric (just right)?

Choose one:
A. Fuel-rich
B. Fuel-lean
C. Stoichiometric

Answer 1

II) Air is 21.0% (by volume) O2.

So, O2 in 24 volume of air ( 24× 21.0 / 100 ) vol.

                                             =5.04 volume

                                             

For all gases, the volume that propane/ oxygen or any other gas occupies is directly proportional to the number of moles of respective gas.

So, we can say, 24 volume of air =( 24×n ) mole of air.

Where "n" is a constant

And, number of moles of O2 gas in 24 volume of air = (5.04×n) mole

(24×n) mole of air contain (5.04×n) mole Oxygen

1 mole of air contain (5.04/24) mole Oxygen

24 mole of air contain 5.04 mole of Oxygen

Combustion of propane by oxygen occurs according to the following equation

C3H8 (g) + 5O2 (g) ------> 3CO2 (g) + 4H2O(g)

So,

 5 moles of Oxygen can combust 1 mole of propane

1 mole of Oxygen can combust (1/5) mole of propane

5.04 mole of Oxygen can combust 5.04/5 mole propane

                                          =1.008 mole propane

5.04 mole of oxygen present in 24 mole of air

Hence,

1.008 moles of propane are needed to completely use up the oxygen in 24.0 moles of air.    

  

I)

1 volume of propen = (1×n) mole of propen

And 24 volume of air contain (24×n) mole air

From the previous calculations,

For every 24 moles of air, 1.008 mole of propane have to present for the gas to become a stoichiometric one.

But in the supplied gas ,

For every 24 moles of air, 1 mole of propane present .

As the amount of propen in the supplied gas is low so, the flame produced by the burner is B. Fuel-lean