# How do you determine the gram-formula mass of the propane gas? State the mole ratio represented in the equation of oxygen to carbon dioxide. Calculate the number of liters of water vapor produced when 25.0 liters of oxygen gas are consumed?

## The burning of propane gas can be represented as a balanced chemical reaction as follows: ${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \to 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(g\right)$

Feb 9, 2017

The gram-molecular weight of any compound is the sum of the atomic weights of each element. The mole ratio in an equation is the ratio of the coefficients of the compounds in a balanced equation. 0.0016L of liquid water are produced, or 19.9L if it all remains in the vapor phase at NTP.

#### Explanation:

For propane we have 3 carbons and 8 hydrogen atoms, so the gram-molecular weight is (312 + 81) = 44g/mol. The mole ratio of oxygen to carbon dioxide is 5:3, so for every five moles of oxygen reacted, three moles of carbon dioxide are produced.

25.0L of oxygen is 25.0/22.4L/mol = 1.12moles of oxygen. The ratio of oxygen to water in the equation is 5:4. Therefore, it will produce

1.12" Moles oxygen" * (4"mol Water")/(5"mol oxygen") = 0.89 moles of water.

0.89 moles of water multiplied by its gram-molecular weight of 18 is 16.1 grams. At a density of 1g/mL, this will be 16.1mL of liquid water, or 0.0016L. If the water is assumed to remain in vapor form, the volume would be 0.89mole * 22.4L/mol = 19.9L