Before reaction
P=730torr= 97332.36 Pa
V = 1L= 0.001 m^3
T= 25 C= 298K
nCO2=0.039 mol
After reaction
P=150torr=20000 Pa
T and V same
nCO2 = 0.008 mol
CO2 which reacted its nCO2 before reaction minus nCO2 after reaction
nCO2 = 0.031 mol
Let be x it's mol of CO2 which reacted with CaO, and y it's mol of which reacted with BaO. We have system of equation
x+y = 0.031
x*MCaO + y*MBaO = 4
When we solved this system x = 0.008
Then m CaO = 0.008* 56 = 0.449 g
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