1st calculate the mol of CO2 reacted.
Given:
P = 258.0 mm Hg
= (258.0/760) atm
= 0.3395 atm
V = 625.0 L
T = 27.0 oC
= (27.0+273) K
= 300 K
find number of moles using:
P * V = n*R*T
0.3395 atm * 625 L = n * 0.08206 atm.L/mol.K * 300 K
n = 8.619 mol
From reaction,
Mol of LiOH required = 2*mol of CO2 required
= 2*8.619 mol
= 17.238 mol
Use:
Mass of LiOH = mol of LiOH * molar mass of LiOH
= 17.238 mol * 23.95 g/mol
= 413 g
Answer: 413 g
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