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

Oct 6, 2016

#### Answer:

A stoichiometrically balanced equation is an absolute prerequisite.

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(g\right)$

#### Explanation:

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(g\right)$

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, $44 \cdot g$ of carbon reactant and $160 \cdot g$ oxidant, $132 \cdot g$ of $C {O}_{2}$ gas are produced along with $72 \cdot g$ 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 $0.75 \cdot m o l$ of propane, coreactant ${O}_{2}$, and product $C {O}_{2}$ are scaled down proportionally. That is, we require $0.75 \cdot m o l \times 5 \cdot m o l$ dioxygen gas, and $0.75 \cdot m o l \times 3$ carbon dioxide gas are evolved.

And thus $C {O}_{2}$ evolved, $=$ $0.75 \cdot m o l \times 3$ $=$ $2.25 \cdot m o l$ $C {O}_{2}$ 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.