I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
Report (any type)/Brief report
22/09/2021
The easiest way is to remember that in order to use , you need to know the moles of the two substances concerned.
We can use the to help us to determine the effect of temperature, pressure, and volume on the number of moles of a gas.
The central requirement of any stoichiometry problem is to convert moles of ##”A”## to moles of ##”B”##.
If ##”A” ##and/or ##”B”## are solids or liquids, you use the mass and molar mass to get moles.
If ##”A”## and/or ##”B”## are gases, you use the to get moles.
Here’s a flow chart to help you through the process.
(From homepage.usask.ca)
EXAMPLE 1
What volume of oxygen at STP is produced when 10.0 g of potassium chlorate decomposes to form potassium chloride and oxygen?
Solution
Let’s see how this works.
Step 1. Write the balanced equation.
##”2KClO”_3 → “2KCl” + “3O”_2##
Step 2. Calculate the moles of ##”KClO”_3##.
##10.0 cancel(“g KClO₃”) × (“1 mol KClO”_3)/(122.6 cancel(“g KClO₃”)) = “0.08156 mol KClO”_3##
Step 3. Calculate the moles of ##”O”_2##.
The balanced equation tells us that 2 mol ##”KClO”_3## give 3 mol ##”O”_2##. Therefore,
##0.08156 cancel(“mol KClO₃”) × (“3 mol O”_2)/(2 cancel(“mol KClO₃”)) = “0.1223 mol O”_2##
Step 4. Convert moles of ##”O”_2## to litres of ##”O”_2##.
Since 1997, STP has been defined as 0 °C and 100 kPa.
The molar volume of an ideal gas at STP is 22.711 L.
At STP, we use the relation 22.711 L = 1 mol. Therefore
##0.1223 cancel(“mol O₂”) × (“22.711 L O”_2)/(1 cancel(“mol O₂”)) = “2.78 L O”_2##
Notice how we always write the conversion factors so that the units cancel to give the desired units for the answer.
If the question asks you to find the volume of gas at some other temperature or pressure, you can use the Ideal Gas Law,
##PV = nRT##.
Suppose the question had asked for the volume at 1.05 atm and 25 °C (298 K). You would write
##V = (nRT)/P = (0.122 cancel(“mol”) × “0.082 06 L·”cancel(“atm·K⁻¹mol⁻¹”) × 298 cancel(“K”))/(1.05 cancel(“atm”)) = “2.85 L”##
EXAMPLE 2
What mass of potassium chlorate is required to produce 3.00 L of oxygen at STP?
Solution
Step 1. The balanced equation is
##”2KClO”_3 → “2KCl” + “3O”_2##
Step 2. Convert litres at STP to moles.
##3.00 cancel(“L O₂”) × (“1 mol O”_2)/(22.711 cancel(“L O₂”)) = “0.1321 mol O”_2##
Step 3. Convert moles of ##”O”_2## to moles of ##”KClO”_3##.
##0.1321 cancel(“mol O₂”) × (“2 mol KClO”_3)/(3 cancel(“mol O₂”)) = “0.088 06 mol KClO”_3##
Step 4. Calculate the moles of ##”KClO”_3##.
##0.08806 cancel(“mol KClO₃”) × (“122.6 g KClO”_3)/(1 cancel(“mol KClO₃”)) = “10.8 g KClO”_3##
If the question asks you to find the volume of gas at some other temperature or pressure, you can use the Ideal Gas Law, ##PV = nRT##.
Suppose the question gave you the volume at 1.05 atm and 25°C (298 K). You would write
##n = (PV)/(RT) = (1.05 cancel(“atm”) × 3.00 cancel(“L”))/(0.082 06 cancel(“L·atm·K⁻¹”)”mol”^-1 × 298 cancel(“K”)) = “0.129 mol O”_2##
Now that you have the moles of ##”O”_2##, you can continue with steps 3 and 4 above.
EXAMPLE 3
Ethylene gas burns in air according to the following equation.
##”C”_2″H”_4″(g)” + “3O”_2″(g)” → “2CO”_2″(g)” + “2H”_2″O(l) “##
If 13.8 L of ##”C”_2″H”_4## at 21°C and 1.083 atm burns completely in oxygen, calculate the volume of ##”CO”_2## produced, assuming the ##”CO”_2##is measured at 44°C and 0.989 atm.
Solution
This one requires a little more work, because you have to use the Ideal Gas Law at the beginning and at the end.
Let’s see how this works. The balanced equation is
##”C”_2″H”_4″(g) “+ “3O”_2″(g)” → “2CO”_2″(g)” + “2H”_2″O(l)”##
Step 1. Calculate the moles of ##”C”_2″H”_4##.
##PV = nRT##
##n = (PV)/(RT) = (1.083 cancel(“atm”) × 13.80 cancel(“L”))/(0.082 06 cancel(“L·atm·K⁻¹”)”mol”^-1 × 294 cancel(“K”)) = “0.619 mol C”_2″H”_4##
Step 2. Calculate the moles of ##”CO”_2##
##0.619 cancel(“mol C₂H₄”) × (“2 mol CO”_2)/(1 cancel(“mol C₂H₄”)) = “1.24 mol CO”_2##
Step 3. Calculate the new volume.
##PV = nRT##
##V = (nRT)/P = (1.24 cancel(“mol”) × “0.082 06 L·”cancel(“atm·K⁻¹mol⁻¹”) × 317 cancel(“K”))/(0.989 cancel(“atm”)) = “32.6 L”##
Here’s a great video showint the relation between stoichiometry and thr Ideal Gas Law.
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
Delivering a high-quality product at a reasonable price is not enough anymore.
That’s why we have developed 5 beneficial guarantees that will make your experience with our service enjoyable, easy, and safe.
You have to be 100% sure of the quality of your product to give a money-back guarantee. This describes us perfectly. Make sure that this guarantee is totally transparent.
Read moreEach paper is composed from scratch, according to your instructions. It is then checked by our plagiarism-detection software. There is no gap where plagiarism could squeeze in.
Read moreThanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result.
Read moreYour email is safe, as we store it according to international data protection rules. Your bank details are secure, as we use only reliable payment systems.
Read moreBy sending us your money, you buy the service we provide. Check out our terms and conditions if you prefer business talks to be laid out in official language.
Read more