What is the molar quantity of carbon dioxide if #0.75*mol# propane gas is combusted completely?

1 Answer
Oct 6, 2016

Answer:

A stoichiometrically balanced equation is an absolute prerequisite.

#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(g)#

Explanation:

#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(g)#

Is this equation stoichiometrically balanced? That is, for every reactant particle, is there a corresponding product particle? In fact there is, and there must be. You might think this is a bit unfamiliar, but if you were receiving change from a cash purchase, you would be able to tell whether you were being short-changed immediately.

Given this, the equation unequivocally tells us that for each mole of propane, #44*g# of carbon reactant and #160*g# oxidant, #132*g# of #CO_2# gas are produced along with #72*g# of water. Mass of course is conserved as in every chemical reaction ever observed.

The equation as written was for 1 mole of propane. If there are #0.75*mol# of propane, coreactant #O_2#, and product #CO_2# are scaled down proportionally. That is, we require #0.75*molxx5*mol# dioxygen gas, and #0.75*molxx3# carbon dioxide gas are evolved.

And thus #CO_2# evolved, #=# #0.75*molxx3# #=# #2.25*mol# #CO_2# gas. What is the corresponding mass of this molar quantity? Can you follow this? If no, voice your objection, and we can go over this. The problem illustrates a fundamental issue of chemical understanding.

See here for another problem illustrating conservation of mass.