i have always been intimidated by Stoichiometry. Now,i decided to start freshly in order to get an A this time. Your channel is very helpful. Thank you so much. I have learned a lot from your videos. :)
Professor Organic Chemistry Tutor, thank you for an outstanding explanation of Gas Stoichiometry Problems at STP in AP/General Chemistry. In Chemistry, conversions are tied to Stoichiometry problems from start to finish. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
Small reminder: He's using the ideal gas constant for ATM and L. That is why it's 0.082057366080960 L⋅atm⋅K−1⋅mol−1 and not 8.31446261815324 J⋅K−1⋅mol−1. Since the 2019 redefinition of SI base units, both NA and k are defined with exact numerical values when expressed in SI units.
**NOTE** For the C3H8 problem, do you think it's important to mention that you can convert directly from Liters to Mole/Mole ratio ONLY if Temperature and Pressure are constant? I'm assuming that you are assuming that T&P are constant. If the problem doesn't mention that T&P are constant, then you have to use Ideal Gas Law twice.
I never pay attention in class, and whenever I try to, it makes me sleepy and I don't learn anything lmao, but with ur videos I feel like I could learn more and manage my time better of how compressed they are compared to 1 hr of class lmao, thx for helping me survive chem this sem. God Bless you TOCT.
Just asking but isn’t it 245 mL of HCl not 245 L, so wouldn’t that affect the number of moles you get for HCl which changes things? Just a concept check
why do you use the ideal gas equation when solving for the mg3n2 equation? why can't we just say that 4.7L of ammonia is equal to the moles as in 4.7mol since the moles are equal to litres in gases please help
Can u help? I have a problem like the last two example. What if, after the reaction, we were asked to find the pressure but is also said that we should neglect the volume of the product. Should I just remove the V from the formula?
"One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 325nm. a) What is the energy of a photon of this wavelength? b) What is the energy of a mole of these photons? c) How many photons are in a 1.00mJ burst of this radiation? d) These UV photons can break chemical bonds in your skin to cause sunburn, a form of radiation damage. If the 325nm radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in kJ/mol."
Can you explain these types of questions please? I tried doing them and ending up getting the right answers but it's only because I kept plugging in numbers to try it out but I still don't understand the concept or what the original equation should be
These questions are the reason I hate chemistry. I would never need to know this information, I would just wear sunscreen. A bunch of fooey if you ask me.
Just gonna go ahead and point out that Standard Pressure is no longer expressed in atm. As of 1982 STP is defined by IUPAC as exactly 1.00 bar , 273.15 K. As a result 22.4 L / 1 mol is no longer valid. Since the standard unit is bar and not atm 22.4 L becomes 22.7. So, 1 mol of gas at STP = 22.7 L
We can use the direct current from Dynamo to hydrogen making and reuse the hydrogen again and again so we go further in lesser fuel Please do the math over it
The complete combustion of a tank of petrol produced 219 kg of carbon dioxide gas: a. What volume of carbon dioxide is produced when measured at STP? b. What mass of oxygen is used by a car in burning a tank of petrol? You may assume that for every 1.00 L of carbon dioxide produced, 1.00 L carbon dioxide produced, 1.00 L of oxygen is consumed. pls help me with this guys!!!!
I may be a little late, but you need to use moles to solve these problems. For a., convert 219kg of into mols of C02, which is approximately 4976.141786 mols. You then use PV=nRT by plugging in the values at STP and the amount of mols you have, solving for V. this will look like: V=((4976.1418)(0.08206)(273))/1. This will equal: 111477.4192L, accounting for significant figures you should get: 111,000L CO2. For b, I'm not sure that this problem is solvable. You would need to know the chemical reaction, and whether or not you can assume STP. If there is no more information then possibly: LCO2=LO2 will be an acceptable answer. For practice I will assume the chemical reaction is gaseous Octane reacting with excess 02 to produce carbon dioxide and water. The balanced chemical reaction should go as follows: C8H12(g) + 11O2(g) -> 8CO2(g) + 6H2O(g) I will also assume the gasses are at STP and the volume of the petrol tank is 50.0L, and I will be solving for the mass of CO2 produced. First we need to solve for the mols of Octane, to do this you need to plug in the information stated with the equation n=(PV)/(RT). This should look like: n= ((1)(50))/((0.08206)(273)). n = 2.2319mol Octane. now multiply the mol ratio of Octane to CO2, which is 8/1 by the mols of Octane we have, which is: 8*2.2319= 12.85524mol CO2. Now we can convert the mols into grams by doing 17.85524mol CO2 * 44.01g/mol CO2 =785.809g CO2, applying significant figures the answer should be: 786g CO2. For a challenge lets assume that the combustion reaction of 1mol of Octane takes place in an enclosed chamber of 1.00L at 100.C which is being pumped with excess O2, which will then be removed when the reaction is finished. What would the pressure of the reacted compounds in the chamber be? to solve this we need to divide mol ratio of Octane used and Octane in the reaction (this is 1/1, if the coefficient of Octane was, for example: 4, and there was still 1 mol of octane actually reacted then the ratio would be 4/4. this removes the coefficient Infront of Octane, I'm not sure how to explain this any better so hopefully this will do). You then divide the products mol ratios by the amount divided by the Octane. This is a really bad way of explaining it so hopefully the view of the balanced chemical equation will help: 1/1C8H12 + 11O2 -> 8/1CO2 + 6/1 H2O. you then take the products mols and plug it into PV=nRT, solving for P. This should look like: P = ((8+6)(0.08206)(373))/1 = 428.5173atm produced. Accounting for significant figures you should get 429atm produced. For b I should should clarify that the problems I did I did not derive from the information given, I simply took the premise to show different ways that you can use gas stoichiometry to solve problems. it would not be possible to derive these problems from the information given, so I made up some information to be able to do the problems I did. I know that I am late to answering you question, which I did so more to help myself in understand gas stichometry then to teach it, so I hope that you now have an answer and understand you original question. I also hope you at least partially understood my explanations. I should also clarify that the problems I did are most likely harder than what you would see on any test of yours, I am in AP chem, and that is were I derived these problems from, so if they felt way too hard then do not worry too much. If you take one thing away from this response then it should be that moles are the key to stoichiometry, and chemistry at that. If you can familiarize yourself with moles then you will have familiarized yourself with stoichiometry. Good luck on your future chemistry endeavors.
We can printed the biological molecules by using an atom a day so we got living molecules in few days and artificial petrol fact9 become possible for whole universe
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You are seriously the best teacher ever. Went from a 62 to an 88 in the span of one month because of you!❤❤
congrats bro
i have always been intimidated by Stoichiometry. Now,i decided to start freshly in order to get an A this time. Your channel is very helpful. Thank you so much. I have learned a lot from your videos. :)
Thanks
@@TheOrganicChemistryTutor You are awesome! :)
@@TheOrganicChemistryTutor I love you
@@TheOrganicChemistryTutor You are awesome!
@@TheOrganicChemistryTutor You are awesome!
Im failing chem and this helped me out alot to understand everything we've learned. Thank you so much.
@Deangelo Duncan Its so obvious this is a scam this is ridiculous
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@@scraic1356 The person I replied to deleted their comment. Your 7 months to late. They were promoting an app that was a scam.
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You were meant to teach and I’m so glad you do. Chemistry has never been clearer to me.
Professor Organic Chemistry Tutor, thank you for an outstanding explanation of Gas Stoichiometry Problems at STP in AP/General Chemistry. In Chemistry, conversions are tied to Stoichiometry problems from start to finish. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
you have the most soothing voice its the only thing pushing me forward in chem
Small reminder: He's using the ideal gas constant for ATM and L. That is why it's 0.082057366080960 L⋅atm⋅K−1⋅mol−1 and not 8.31446261815324 J⋅K−1⋅mol−1. Since the 2019 redefinition of SI base units, both NA and k are defined with exact numerical values when expressed in SI units.
🤓
@@zetashangu don't let me see you comment on this youtube page or else it's over for your baboon like face who still wear facemasks 🙏
Thank you :)
no one cares and this is not relevent for basic stoichiometry. nerd.
@@zetashanguthis is why ur here watching a chemistry practice video
the thing i love about you is you always come up with different examples and sol thanks a lot for being here to help us out
Thank you! Appreciate you doing it slow and in an understanding way.
**NOTE** For the C3H8 problem, do you think it's important to mention that you can convert directly from Liters to Mole/Mole ratio ONLY if Temperature and Pressure are constant? I'm assuming that you are assuming that T&P are constant. If the problem doesn't mention that T&P are constant, then you have to use Ideal Gas Law twice.
If no change in T or P is stated or hinted at then they are assumed to be constant.
Idk
You are the best, keep the work you are saving everyone's grades
thanks to you i passed my chemistry exam, IN FACT I DID BETTER THAN I THOUGHT, thanks man
Happy for u
This helped so much thank you... my chem teacher doesn't know how to teach so I use u to learn
This is so helpful. I can't understand our lesson but after watching this I able to understand everything. Thank you so much🥰
I learn infinitely better from you than the nonexistent teachings of my teacher.
I never pay attention in class, and whenever I try to, it makes me sleepy and I don't learn anything lmao, but with ur videos I feel like I could learn more and manage my time better of how compressed they are compared to 1 hr of class lmao, thx for helping me survive chem this sem. God Bless you TOCT.
YOOOO this is exactly what I needed and I didn't even search for this video. This is amazing thank you.
Very helpful vid! Understood it better than when my teacher taught it. :)
Thank you so much we are learning a lot .I was blank on these .it's now clear
I have a hard time learning during lecture but this guy makes all topics incredibly easy to learn.
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Congratulations 1M subscriber
This has been really helpful thanks a lot
thank you so very much! i fell asleep during the introduction of gas stoich in my chem class and now im super lost, haha...
thanks so much, this stumped me
Thank u so much ur so better than my gen chem teacher
this really helps me a lot
Can you just get the moles of argon by saying if one mole is 22.4dm^3 then 2.5 moles of argon are 2.5 times 22.4 which is 56dm^3 thanks😊😊
This is so helpful! Thank you🌷
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Congrats for 1M subscribers
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@@راكانالغامدي-و1عwhat is bro yappin abt
PV=nRT
U can also use this
For the first question....
My class use Kpa and not atm. Would that affect anything? For example in the first equation in the video.
I need to study becauce im failing chem with this topic, thank you for shredding some light because i'm on the tip of giving up.
So helpful! Thank you!
Just asking but isn’t it 245 mL of HCl not 245 L, so wouldn’t that affect the number of moles you get for HCl which changes things? Just a concept check
You have to have your answer in L, not mL so, you would have to do the conversion from mL, to L. Which is what he did.
Your channel is very helpful thank you so much I got 95 on the test after I watch this
good for you
the 1 mole of gas = 22.4L is that true for all gasses or just argon?
ALL GASES
@@corecube7721 dope thanks lol stupid question ik thanks again
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why do you use the ideal gas equation when solving for the mg3n2 equation? why can't we just say that 4.7L of ammonia is equal to the moles as in 4.7mol since the moles are equal to litres in gases please help
Can u help? I have a problem like the last two example. What if, after the reaction, we were asked to find the pressure but is also said that we should neglect the volume of the product. Should I just remove the V from the formula?
"One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 325nm. a) What is the energy of a photon of this wavelength? b) What is the energy of a mole of these photons? c) How many photons are in a 1.00mJ burst of this radiation? d) These UV photons can break chemical bonds in your skin to cause sunburn, a form of radiation damage. If the 325nm radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in kJ/mol."
Can you explain these types of questions please? I tried doing them and ending up getting the right answers but it's only because I kept plugging in numbers to try it out but I still don't understand the concept or what the original equation should be
These questions are the reason I hate chemistry. I would never need to know this information, I would just wear sunscreen. A bunch of fooey if you ask me.
Thats a tough question is that higher level chemistry?!?!?!?
U R MY HERO!!
This chemistry video tutorial explains how to solve gas stoichiometry practice problems at STP and not at STP. Did you mean STAP for the last one?
24:10 isn’t 0.0821?
thank you a again all though it was kind of tough i made it work.
but i think that formula is V=n*24 is it wrong
wow good job
thank you so much
My teacher literally didn’t teach us that 1 mol = 22.4L so I’ve been sitting here wondering how I have 1487.39582824 mol of N2…….
Just gonna go ahead and point out that Standard Pressure is no longer expressed in atm. As of 1982 STP is defined by IUPAC as exactly 1.00 bar , 273.15 K. As a result 22.4 L / 1 mol is no longer valid. Since the standard unit is bar and not atm 22.4 L becomes 22.7. So, 1 mol of gas at STP = 22.7 L
tf man
i dunno, mcat still uses 22.4 but the numbers are so similar, pretty sure the difference is negligible unless you're a physical chemist..
Thank you
i don't understand the 6th problem, how come HCl is the limiting when Zn has the lesser value of moles?
@@thomaspaulson6202 ah yeah thank you:) I seem to have overlooked that part:)
How
God bless you!
Short cuts tend to not help people. I would strongly suggest using Idea Gas Laws and staying consistent in your methods vs. taking shortcuts.
for #2 at 6:36
We can use the direct current from Dynamo to hydrogen making and reuse the hydrogen again and again so we go further in lesser fuel
Please do the math over it
thx babe ❤❤
The complete combustion of a tank of petrol produced 219 kg of carbon dioxide gas:
a. What volume of carbon dioxide is produced when measured at STP?
b. What mass of oxygen is used by a car in burning a tank of petrol? You may assume that
for every 1.00 L of carbon dioxide produced, 1.00 L carbon dioxide produced, 1.00 L of
oxygen is consumed.
pls help me with this guys!!!!
I may be a little late, but you need to use moles to solve these problems. For a., convert 219kg of into mols of C02, which is approximately 4976.141786 mols. You then use PV=nRT by plugging in the values at STP and the amount of mols you have, solving for V. this will look like: V=((4976.1418)(0.08206)(273))/1. This will equal: 111477.4192L, accounting for significant figures you should get: 111,000L CO2.
For b, I'm not sure that this problem is solvable. You would need to know the chemical reaction, and whether or not you can assume STP. If there is no more information then possibly: LCO2=LO2 will be an acceptable answer. For practice I will assume the chemical reaction is gaseous Octane reacting with excess 02 to produce carbon dioxide and water. The balanced chemical reaction should go as follows: C8H12(g) + 11O2(g) -> 8CO2(g) + 6H2O(g) I will also assume the gasses are at STP and the volume of the petrol tank is 50.0L, and I will be solving for the mass of CO2 produced. First we need to solve for the mols of Octane, to do this you need to plug in the information stated with the equation n=(PV)/(RT).
This should look like: n= ((1)(50))/((0.08206)(273)). n = 2.2319mol Octane. now multiply the mol ratio of Octane to CO2, which is 8/1 by the mols of Octane we have, which is:
8*2.2319= 12.85524mol CO2. Now we can convert the mols into grams by doing 17.85524mol CO2 * 44.01g/mol CO2 =785.809g CO2, applying significant figures the answer should be: 786g CO2. For a challenge lets assume that the combustion reaction of 1mol of Octane takes place in an enclosed chamber of 1.00L at 100.C which is being pumped with excess O2, which will then be removed when the reaction is finished. What would the pressure of the reacted compounds in the chamber be? to solve this we need to divide mol ratio of Octane used and Octane in the reaction (this is 1/1, if the coefficient of Octane was, for example: 4, and there was still 1 mol of octane actually reacted then the ratio would be 4/4. this removes the coefficient Infront of Octane, I'm not sure how to explain this any better so hopefully this will do). You then divide the products mol ratios by the amount divided by the Octane. This is a really bad way of explaining it so hopefully the view of the balanced chemical equation will help: 1/1C8H12 + 11O2 -> 8/1CO2 + 6/1 H2O. you then take the products mols and plug it into PV=nRT, solving for P. This should look like: P = ((8+6)(0.08206)(373))/1 = 428.5173atm produced. Accounting for significant figures you should get 429atm produced.
For b I should should clarify that the problems I did I did not derive from the information given, I simply took the premise to show different ways that you can use gas stoichiometry to solve problems. it would not be possible to derive these problems from the information given, so I made up some information to be able to do the problems I did. I know that I am late to answering you question, which I did so more to help myself in understand gas stichometry then to teach it, so I hope that you now have an answer and understand you original question. I also hope you at least partially understood my explanations. I should also clarify that the problems I did are most likely harder than what you would see on any test of yours, I am in AP chem, and that is were I derived these problems from, so if they felt way too hard then do not worry too much. If you take one thing away from this response then it should be that moles are the key to stoichiometry, and chemistry at that. If you can familiarize yourself with moles then you will have familiarized yourself with stoichiometry. Good luck on your future chemistry endeavors.
@@Rotnoc473 Thank you so much... 🥺
this doesnt work, becuase for PV= nRT doesnt the pressure have to be in kpa, which would make it 101.3 not 1 atm
What is the volume of 7.6g of argon at RTP? guys I'm failing tomorrow 😭
We can printed the biological molecules by using an atom a day so we got living molecules in few days and artificial petrol fact9 become possible for whole universe
Sobbinggggggg😭😭
Finals tommorow : (
Im pretty sure number 1 is wrong because its supposed to be2.5 mols but he puts 1 mold and also he didint solve the whole equation for number 1
5 + 5 = 55
yes
yes
2020 people
the way none of these set ups for the problems got me the right answers for my chemistry worksheets….do better thanks !
if you were carti, it would be easier to understand, but aight bruh
BENETTE
help
Stem pa more
i hatte this
The answer you gave to the first problem is wrong.You needed to plug the 56L into the problem to get the actual answer of 1.000593
Kenny Jenkins
I think u r wrong
shut up nerd
you don't explain well
Gggrrrrr... I'm not interested
Congratulations 1M subscriber
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Thank you