Question

One method of sequestering CO2 in spacecraft is allowing the CO2 to react with LIOH to produce LizCO, and H20. CO2(g) + 2 LiOH (s) — LizCo (s) + H20 (8) If 30.0 g of LiOH is allowed to react with excess CO2, what is the theoretical yield in grams of Li2CO3 that would be formed? After the above experiment with 30,0 g of LiOH has been performed, a technician determines the mass of the LizC03 product to be 25.4 grams. What is the percent yield for the reaction?

Answer 1

1)

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 = 30 g

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

= 30/23.98

= 1.251 mol

According to balanced equation

mol of Li2CO3 formed = (1/2)* moles of LiOH

= (1/2)*1.251

= 0.6256 mol

Molar mass of Li2CO3,

MM = 2*MM(Li) + 1*MM(C) + 3*MM(O)

= 2*6.968 + 1*12.01 + 3*16.0

= 73.946 g/mol

mass of Li2CO3 = number of mol * molar mass

= 0.6256*73.95

= 46.26 g

Answer: 46.3 g

2)

% yield = actual mass*100/theoretical mass

= 25.4*100/46.26

= 54.9 %

Answer: 54.9 %