2C2H2 + 5O2 ==> 4CO2 + 2H2O
Use PV = nRT to solve for moles C2H2.
Use PV = nRT to solve for moles O2.
Then substitute these using the worked example in this link to solve for moles CO2 and mole H2O produced.
http://www.jiskha.com/science/chemistry/stoichiometry.html
1. A 123.00 liter sample of acetylene gas (C2H2) is reacted with excess oxygen at 27 C and 1 atm to produce carbon dioxide and water vapor. Calculate the following:
a. The volume of CO2 produced at 1 atm and 400 C.
b. The mass of H2O vapor produced.
3 answers
ok. So I already had the balanced equation. using PV=nRT is just confusing me. I am left with 749 K for C2H2 and 123.15 as V for O2. How am I supposed to use those values for finding the volume of CO2 produced at 1 atm an 400 C? I asked the question because I need it to be broken down for me-NOT so I can be referred to an explanation that isn't directed towards my problem and is even more confusing.
2C2H2 + 5O2 ==> 4CO2 + 4H2O
In chemistry, moles react with moles. That's why you must convert 123.00 L C2H2 at the conditions listed to determine how many moles you have.
PV = nRT is the only way I know how to do it. If you know a better way please let me in on the secret, too.
n = PV/RT = 1 atm x 123L/(0.08206)(273+27) and n = approximately 5 moles (but you need to go through the calculation and confirm that. That isn't the EXACT value.)
If you had gone through the link I gave you (which by the way will work ALL of your stoichiometry problems) you would have seen how to solve stoichiometry problems.
Convert moles C2H2 to moles CO2 using the coefficients in the balanced equation.
5 moles C2H2 x (4 moles CO2/2 moles C2H2) = 5 x (4/2) = 10 moles CO2 (produced at STP). You want the volume produced at 400 C and 1 atm. So use
PV = nRT, you have P = 1 atm, V = ?, n = 10 moles, R = 0.08206 L*atm and T = 400 C which is 673 K. {By the way I apologize for the typo in my original response; it should have read moles CO2 instead of moles O2.)
All of this gives you part a in your post which is volume of CO2 produced by 123 L C2H2(at the conditions listed) and volume of CO2 at 1 atm and 400 C. The answer you should get is about 550 L (again, not an exact number).
Part b for mass H2O is done the same way, using the above as a guide or using the link I gave you earlier. I'll be glad to help you through it IF you show your work AND explain fully what you don't understand about the next step.
In chemistry, moles react with moles. That's why you must convert 123.00 L C2H2 at the conditions listed to determine how many moles you have.
PV = nRT is the only way I know how to do it. If you know a better way please let me in on the secret, too.
n = PV/RT = 1 atm x 123L/(0.08206)(273+27) and n = approximately 5 moles (but you need to go through the calculation and confirm that. That isn't the EXACT value.)
If you had gone through the link I gave you (which by the way will work ALL of your stoichiometry problems) you would have seen how to solve stoichiometry problems.
Convert moles C2H2 to moles CO2 using the coefficients in the balanced equation.
5 moles C2H2 x (4 moles CO2/2 moles C2H2) = 5 x (4/2) = 10 moles CO2 (produced at STP). You want the volume produced at 400 C and 1 atm. So use
PV = nRT, you have P = 1 atm, V = ?, n = 10 moles, R = 0.08206 L*atm and T = 400 C which is 673 K. {By the way I apologize for the typo in my original response; it should have read moles CO2 instead of moles O2.)
All of this gives you part a in your post which is volume of CO2 produced by 123 L C2H2(at the conditions listed) and volume of CO2 at 1 atm and 400 C. The answer you should get is about 550 L (again, not an exact number).
Part b for mass H2O is done the same way, using the above as a guide or using the link I gave you earlier. I'll be glad to help you through it IF you show your work AND explain fully what you don't understand about the next step.
3 answers