Propane, C 3 H 8 , burns in molecular oxygen, O 2 , to give water, H 2 O, and ca
ID: 1009285 • Letter: P
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) + 5 O2(g) =====> 3 CO2(g) + 4 H2O(g)
Temperature we can fix 273 K and pressure is 1 atm
PV = nRT
1 x 5.2 = n X 0.0821 x 273
n = 4.31 Mole
4.31 Mole of propane (190.05 g) needs 21.55 Mole of oxygen (5 equivalents)
21.55 Mole of oxygen = 21.55 x 32 = 689.574 gm
PV = nRT
V x 1 = 21.55 x 0.0821 x 273
V = 483 Liter of oxygen is need
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