Question

Lithium Hydroxide 2LiOH+CO2, what is the volume of CO2 at 23°C and 777mmHg that could be absorved by 1 lb of LiOH?

Answer 1

mass of LiOH = 1 lb

= 453.6 g

Molar mass of LiOH,

MM = 1*MM(Li) + 1*MM(O) + 1*MM(H)

= 1*6.968 + 1*16.0 + 1*1.008

= 23.976 g/mol

mass of LiOH = 4.536*10^2 g

mol of LiOH = (mass)/(molar mass)

= 4.536*10^2/23.98

= 18.92 mol

According to balanced equation

mol of CO2 required = (1/2)* moles of LiOH

= (1/2)*18.92

= 9.459 mol

Given:

P = 777.0 mm Hg

= (777.0/760) atm

= 1.0224 atm

n = 9.459 mol

T = 23.0 oC

= (23.0+273) K

= 296 K

use:

P * V = n*R*T

1.0224 atm * V = 9.459 mol* 0.08206 atm.L/mol.K * 296 K

V = 225 L

Answer: 225 L