Homework Answers
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 %
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