Question

16. What volume of water vapor gas would be produced from the combustion of 535.55 grams of propane (C3H8) with 2,665.06 grams of oxygen gas, under a pressure of 0.96 atm and a temperature of 350. degrees C? Given:
C3H8(g) + 5 O2(g) ---> 3 CO2(g) + 4 H2O(g)

(OR C3 H8 ("g") + 5 O2 ("g") right arrow 3 C O2 ("g") + 4 H2O ("g")




Do not type units with your answer

Answer 1

The volume of water vapour gas that would be produced will be 2,581.73 L

Stoichiometric calculations

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 535.55 grams of propane = 535.55/44.1 = 12.12 moles

Mole of 2665.06 grams of oxygen = 2665.06/32 = 83.28 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

12.12 x 4 = 48.48 moles.

Using the ideal gas equation: pv = nRt

v = nRT/p = 48.48 x 0.08206 x 623/0.96 = 2,581.73 L

More on ideal gas equations can be found here: brainly.com/question/4147359

#SPJ1