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 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 porduct, with the larger percent composition by mass of oxygen, are produced?
C.) How many molecules of un namedd reactant are consumed?
You didn't provide the kg propane gas.
Use the next post and see the example. Almost the same process.
Sure, I can help you with that! Let's go through each question step by step.
A.) To determine the moles of the product with the smaller percent composition by mass of oxygen, we need to know the balanced chemical equation for the combustion of propane. The balanced equation is as follows:
C3H8 + 5O2 → 3CO2 + 4H2O
From the equation, we can see that for every 3 moles of carbon dioxide (CO2) produced, 5 moles of oxygen (O2) are consumed. This means that the smaller percent composition by mass of oxygen is for carbon dioxide.
To calculate the number of moles of carbon dioxide, we need to know the mass of propane in kilograms. Let's assume we have 1 kg of propane.
The molar mass of propane (C3H8) can be calculated by adding up the atomic masses of each element:
(3 × atomic mass of carbon) + (8 × atomic mass of hydrogen)
Molar mass of carbon: 12.01 g/mol
Molar mass of hydrogen: 1.01 g/mol
Molar mass of propane (C3H8): (3 × 12.01) + (8 × 1.01) = 44.11 g/mol
Now, we can calculate the number of moles of propane in 1 kg:
Number of moles of propane = Mass of propane (in grams) / Molar mass of propane
Number of moles of propane = 1000 g / 44.11 g/mol
Number of moles of propane = 22.68 mol
Since the coefficient ratio between carbon dioxide and propane is 3:1, we can determine that:
Number of moles of carbon dioxide = (3/1) × Number of moles of propane
Number of moles of carbon dioxide = (3/1) × 22.68 mol
Number of moles of carbon dioxide = 68.04 mol
Therefore, 68.04 moles of carbon dioxide are produced.
B.) To determine the grams of the product with the larger percent composition by mass of oxygen (water in this case), we can use the same information from the previous question.
In the balanced equation, it shows that for every 4 moles of water (H2O) produced, 5 moles of oxygen (O2) are consumed. This means that the larger percent composition by mass of oxygen is for water.
To calculate the grams of water, we need to convert the moles of water to grams.
Molar mass of water (H2O): (2 × atomic mass of hydrogen) + (1 × atomic mass of oxygen)
Molar mass of hydrogen: 1.01 g/mol
Molar mass of oxygen: 16.00 g/mol
Molar mass of water (H2O): (2 × 1.01) + (1 × 16.00) = 18.02 g/mol
Number of moles of water = (4/5) × Number of moles of carbon dioxide
Number of moles of water = (4/5) × 68.04 mol
Number of moles of water = 54.43 mol
Now, we can calculate grams of water using the molar mass:
Mass of water = Number of moles of water × Molar mass of water
Mass of water = 54.43 mol × 18.02 g/mol
Mass of water = 980.74 g
Therefore, 980.74 grams of water are produced.
C.) To determine the number of molecules of the unnamed reactant (propane), we need to convert moles of propane to molecules.
We already know that 1 kg of propane is equal to 22.68 moles of propane (from part A).
Avogadro's number states that 1 mole of any substance contains 6.022 x 10^23 molecules. Therefore, we can calculate the number of molecules using this relationship:
Number of molecules of reactant = Number of moles of reactant × Avogadro's number
Number of molecules of propane = 22.68 mol × 6.022 x 10^23 molecules/mol
Number of molecules of propane = 1.365 x 10^25 molecules
Therefore, 1.365 x 10^25 molecules of propane are consumed.
I hope this explanation helps! You can now go back to Bobby and explain the answers to him.