Propane, C3H8, burns in molecular oxygen, O2, to give water, H2O, and carbon dio

Question

Propane, C3H8, burns in molecular oxygen, O2, to give water, H2O, and carbon dioxide, CO2, as shown below

C3H8 (g) + O2 (g) -> CO2 (g) + H2O (g)

For your calculation use the following molar mass for the elements: H (1.01g/mol), C(12.02g/mol), O (16.00g/mol). Assume all gases folow the ideal gas law and are measured at the same temperature and pressure.

a) rewrite equilibrate and balance the above chemical reaction

b) determine the volume in liter (L) of molecular oxygen, O2, required for completely react with 5.2L of C3H8 , Calculate the mass percentage composition in oxygen.

c) What volume in liter (L) of water, H2O vapor produced?

Explanation / Answer

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

b. 1 mole of C3H8 combustion with 5 moles of O2

     22.4 L of C3H8 combustion with 5*22.4L of O2

    5.2L of C3H8 combustion with = 5*22.4*5.2/22.4 = 26L of O2

1 mole of C3H8 combustion with 5 moles of O2

22.4L of C3H8 combustion with 5*32g of O2

5.2L of C3H8 combustion with = 5*32*5.2/22.4 = 37.14g

the mass percentage composition in oxygen = 37.14*100/160 = 23.21%

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

    1 mole of C3H8 combustion with 4 moles of H2O

    22.4L of C3H8 combustion with 4*22.4L of H2O

5.2L of C3H8 combustion with = 4*22.4*5.2/22.4   = 20.8L of H2O

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