Question

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

Answer 1

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.