If 0.45 grams of water is formed during the cumbustion of propane, how many grams of carbon dioxide is formed?

0.825 g of CO2 will be formed.

Explanation:

First of all we should display the reaction of combustion of propane:

C3H8 + 5O2 → 3CO2 + 4H2O

The stoichiometry of this reaction shows that every mole of propane reacts with 5 moles of oxygen to be burned and produce 3 moles of CO2 and 4 moles of water.This mains that every 3 moles of CO2 produced with the production of 4 moles of water.In the problem 0.45 g of water is formed. We can get the number of moles through the relation n = mass/molar mass, mass = 0.45 g and molar mass of water = 18.0 g/mol. n = 0.45/18 = 0.025 mole.As we mentioned that every 3 moles of CO2 produced with 4 moles of water, then we can get the number of moles of CO2 when 0.025 mole of water formed.The number of moles of CO2 corresponding to the formation of 0.025 mole of water = (0.025 * 3) / (4) = 0.01875 mole.Finally we can get the number of grams of CO2 formed; mass = n * molar mass, molar mass of CO2 = 44.0 g/mole.Mass of CO2 = (0.01875 mole * 44.0 g/mole) = 0.825 g.

The kinetic energy of the liquid increases. hope this !

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