Question

How many moles of lithium hydroxide would be required to produce 33.0 g of Li₂CO₃ in the following chemical reaction? 2 LiOH(s) + CO₂ (g) → Li₂CO₃ (s) + H₂O (l)

Answer 1

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 = 33 g
mol of Li2CO3 = (mass)/(molar mass)
= 33/73.95
= 0.4463 mol


According to balanced equation
mol of LiOH required = (2/1)* moles of Li2CO3
= (2/1)*0.4463
= 0.8925 mol

Answer: 0.892 mol