Answer to Question #94137 in General Chemistry for Ben Love

Question

A liquid propane take contains an unknown mass of propane. The propane completely reactions with 23.03 kg of oxygen to form 19.01 kg of carbon dioxide and 10.37 kg of water vapor. What was the unknown mass of propane in the tank before the reaction?

Answer 1

Propane reacts with oxygen by the equation:

C₃H₈ + 5 O₂ = 3 CO₂ + 4 H₂O

Firstly, found the amount in moles of oxygen consumed and CO₂ and H₂O produced:

$n(O_2)=\frac{m(O_2)}{M(O_2)}=\frac{23030 g}{32 \frac{g}{mol}}=720 mol$

$n(CO_2)=\frac{m(CO_2)}{M(CO_2)}=\frac{19010 g}{44 \frac{g}{mol}}=432 mol$

$n(H_2O)=\frac{m(H_2O)}{M(H_2O)}=\frac{10370 g}{18 \frac{g}{mol}}=576 mol$

Now we can check whether oxygen was taken in excess or it's a limiting reagent. According to stoichiometry of the reaction n(O₂):n(CO₂):n(H₂O) = 5:3:4. According to amounts of oxygen consumed and products formed n(O₂):n(CO₂):n(H₂O) = 720:432:576=5:3:4. Thus we can conclude that oxygen has been taken according to stoichiometry.

The amount of propane in moles is equal to

$n(C_3H_8)=\frac{1}{5}n(O_2)=\frac{1}{3}n(CO_2)=\frac{1}{4}n(H_2O)=144mol$

Thus the unknown mass of liquid propane in the tank

$m(C_3H_8)=n(C_3H_8) \cdot M(C_3H_8)=144 mol \cdot 44 \frac{g}{mol}=6336g \approx 6.34 kg$