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?

Md Ariful Islam

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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 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

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