Hank Hill loves everything about gas propane, C3H8. One day his son Bobby asks him the following set of questions about propane. Hank hems and haws for a few moments and finally tells his son he'll get back to him with answers but he has to mow the lawn first. Instead of mowing the lawn, Hank runs up to you and says,"Hey, you're one of those smart chemistry people aren't ya?" What are the answers to Bobby's questions and show me all the work so i can explain it to him."
Bobby asked, if 40.0 kg of propane gas undergoes complete combustion to form carbon dioxide gas and dihydrogenmonoxide gas.
A.) How many moles of the product, with the smaller percent composition by mass of oxygen, are produced?
B.) How many grams of the product, with the larger percent composition by mass of oxygen, are produced?
C.) How many molecules of un namedd reactant are consumed?
See your previous problem. Percent compositions are done the same way. The example link will solve the stoichiometry part of the problem.
A) To determine the number of moles of the product with the smaller percent composition by mass of oxygen, we first need to understand the chemical equation for the combustion of propane:
C3H8 + 5O2 → 3CO2 + 4H2O
From the balanced equation, we can see that for every 5 moles of oxygen (O2) consumed, 3 moles of carbon dioxide (CO2) are produced. Therefore, the smaller percent composition by mass of oxygen will be in carbon dioxide.
The molar mass of carbon dioxide (CO2) is calculated as follows:
C = 12.01 g/mol
O = 16.00 g/mol (2 atoms of oxygen per molecule)
Molar mass of CO2 = 12.01 + (16.00 x 2) = 44.01 g/mol
Now, we can find the number of moles of carbon dioxide produced. First, we need to convert the mass of propane (40.0 kg) to the number of moles of propane.
Molar mass of propane (C3H8) = (3 x 12.01) + (8 x 1.01) = 44.11 g/mol
Number of moles of propane = mass of propane / molar mass of propane
= 40,000 g / 44.11 g/mol
= 907.60 mol
Since the stoichiometric ratio of propane to carbon dioxide is 1:3, we multiply the number of moles of propane by the stoichiometric coefficient of carbon dioxide (3) to find the number of moles of carbon dioxide:
Number of moles of CO2 = number of moles of propane x (3 moles of CO2 / 1 mole of propane)
= 907.60 mol x 3
= 2722.8 mol
Therefore, 2722.8 moles of carbon dioxide (CO2) will be produced.
B) To determine the mass of the product with the larger percent composition by mass of oxygen, we need to consider the ratio between the molar mass of water (H2O) and carbon dioxide (CO2).
Molar mass of water (H2O) = (2 x 1.01) + (16.00) = 18.02 g/mol
Molar mass of CO2 = 44.01 g/mol (as calculated before)
Since the number of moles of water produced is equal to the number of moles of propane consumed, we can calculate the mass of water produced as follows:
Mass of water = number of moles of water x molar mass of water
= 907.60 mol x 18.02 g/mol
= 16,366.9 g
= 16.37 kg (rounded to two decimal places)
Therefore, 16.37 kg of water will be produced.
C) To determine the number of molecules of the unnamed reactant consumed, we need to find the number of moles of the unnamed reactant.
Let's assume the unnamed reactant is oxygen (O2). From the balanced equation, we know that for every 5 moles of oxygen consumed, 3 moles of carbon dioxide are produced. Therefore, the ratio of moles of oxygen to moles of carbon dioxide is 5:3.
Number of moles of oxygen = number of moles of carbon dioxide x (5 moles of oxygen / 3 moles of carbon dioxide)
= 2722.8 mol x (5 / 3)
= 4538.0 mol
Now, to find the number of molecules of oxygen, we need to multiply the number of moles by Avogadro's number (6.022 x 10^23 molecules/mol).
Number of molecules of oxygen = number of moles of oxygen x Avogadro's number
= 4538.0 mol x (6.022 x 10^23 molecules/mol)
= 2.734 x 10^27 molecules
Therefore, approximately 2.734 x 10^27 molecules of oxygen will be consumed.
You can use these explanations and calculations to explain the answers to Bobby's questions.