View previous topic :: View next topic 
Author 
Message 
Caio06
Jr. Member
Joined: 16 Sep 2018
Posts: 5


HELP!!! I need help with this question


A bottle with a volume of 0.85 L is filled up with CO2 at a pressure of 1.44 atm and temperature of 312 K. Upon adding LiOH into this bottle, it is found that the pressure of CO2 reduces to 0.56 atm because some of CO2 reacts with LiOH. Calculate the amount of Li2CO3 that is generated (Li = 7, C = 12,O = 16, H = 1)
CO2(g) + 2LiOH(aq) = Li2CO3(aq) + H2O(l) [/b] 

Back to top 


Caio06
Jr. Member
Joined: 16 Sep 2018
Posts: 5




my results do not correspond to the result given by the teacher 

Back to top 


ChenBeier
Distinguished Member
Joined: 27 Sep 2017
Posts: 287
Location: Berlin, Germany




Show your calculation first.
Do you know ideal gas law?
What are the constant values in this case. What is relation of mole and pressure. 

Back to top 


Caio06
Jr. Member
Joined: 16 Sep 2018
Posts: 5




First i calculated the numbers of moles of CO2 using the first pressure given(1,44atm).
n=PV/RT
n=(1,44)x(0,85)/(0,08206)x(312)
n=0,048 mol of CO2
After i did the same using the last pressure given (0,56 atm) using the same volume ,temperature, which give me 0,0189 moles, so i did the difference between the first and the last what give me 0,029 moles, what i supose is the moles of Li2CO3.
Lastly i did the stoichiometry to find the mass.
0,029mol Li2C03 x (73,9g Li2CO3/1 mol Li2C03)
=2,14 g of Li2CO3
The result given by the teacher is 1,628g. 

Back to top 


ChenBeier
Distinguished Member
Joined: 27 Sep 2017
Posts: 287
Location: Berlin, Germany




Your calculation is correct. I got the same result. Maybe you can get the calculation from your teacher. 

Back to top 


Caio06
Jr. Member
Joined: 16 Sep 2018
Posts: 5


Back to top 


ChenBeier
Distinguished Member
Joined: 27 Sep 2017
Posts: 287
Location: Berlin, Germany




Do you have the teachers calculation? 

Back to top 


Caio06
Jr. Member
Joined: 16 Sep 2018
Posts: 5




Not yet, when I have it, I'll post it here. 

Back to top 


