The balanced combustion reaction of propane, C₃H₈, is
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
Molar mass of propane: 44 g/mol
Moles of propane = 42 g * (1 mol/44g) = 0.9545 mol propane
Molar mass of oxygen: 32 g/mol
Moles of oxygen = 115 g * (1 mol/32 g) = 3.594 mol oxygen
Moles of oxygen needed to completely react propane:
0.9545 mol propane * (5 mol O₂/1 mol propane) = 4.7725 mol oxygen
Since the available oxygen is only 3.594 moles and propane needs 4.7725 moles, that means oxygen is our limiting reactant. We base the amount of water produced here.
Molar mass of water: 18 g/mol
Mass of water produced = 3.594 mol O₂ * (4 mol H₂O/5 mol O₂) * (18 g/mol)
Mass of water produced = 258.768 grams
The percentage composition of CF4 is %C = 13.64% and %F = 86.36%
The solution are as follows:
Molar mass of CF4 = 88
<span>molar mass of C = 12 </span>
<span>molar mass of F = 4x19 = 76 </span>
<span>% C = 12/88 x 100 = 13.64% </span>
<span>% F = 76/88 x 100 = 86.36%</span>
