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]
HELP!!! I need help with this question
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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.
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.