Question

Hi please help me!!

Propane (C3H8) burns in oxygen to form CO2 and H2O according to the following equation. How many grams of O2 are required to burn 2.56 x 1022 propane molecules?

(This chemical equation is not balanced. You need to balance this chemical equation first before calculation)

C3H8 + O2 --> CO2 + H2O

Answer 1

The balanced equation for the combustion of propane is:

C3H8 + 5O2 --> 3CO2 + 4H2O

So for every propane molecule, we need 5 oxygen molecules.

To calculate the number of oxygen molecules required to burn 2.56 x 10^22 propane molecules, we need to multiply the number of propane molecules by the ratio of oxygen molecules to propane molecules.

Ratio of O2 to C3H8 = 5:1

Number of O2 molecules required = (5/1) x 2.56 x 10^22 = 1.28 x 10^23

Now we can convert the number of oxygen molecules to grams using the molar mass of oxygen.

1 mole of O2 = 32 g

1.28 x 10^23 molecules of O2 = (1.28 x 10^23 / 6.022 x 10^23) moles of O2

Mass of O2 = (1.28 x 10^23 / 6.022 x 10^23) x 32 g

Mass of O2 = 6.82 grams

So, 6.82 grams of O2 are required to burn 2.56 x 10^22 propane molecules.