The question requires us to calculate the amount of vapor water produced as a by-product when 66g of carbon dioxide are obtained from the combustion of propane.
Considering the combustion of propane (C3H8), we have the following reaction:
IFrom the reaction, we can see that the stoichiometric relationship between C3H8 and water (H2O) is as follows:
2 mol C3H8 ---------- 8 mol H2O
Then, to calculate the amount of water produced as a by-product, we'll need to determine the amount of reactant needed to produce 66g of CO2.
Since the molar mass of CO2 is 44g/mol and considering the reaction written above, we can write:
2 mol C3H8 ---------- 4 mol CO2
x ---------- (66g/44g) = 1.5 mol CO2
Solving for x, we have that 0.75 mol of C3H8 are required to produce 66g of CO2.
Now, we calculate the amount of water that should be obtained from 0.75 mol of C3H8:
2 mol C3H8 ---------- 8 mol H2O
0.75 mol C3H8 ----- y
Solving for y, we have that 3 moles of water will be obtained as a by-product.
At last, we convert the calculated amount of vapor water into its volume considering the Standard Temperature and Pressure conditions (STP), where 1 mol of a gas corresponds to 22.4 L of the same gas:
1 mol vapor H2O ---------- 22.4 L vapor H2O
3 mol vapor H2O --------- z
Solving for z, we have that 67.2 L of vapor water will be obtained as a by-product when 66g of CO2 are produced from the combustion of propane.