Question

How many moles of lithium hydroxide would be required to produce 32.5 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 = 32.5 g
mol of Li2CO3 = (mass)/(molar mass)
= 32.5/73.95
= 0.4395 mol

According to balanced equation
mol of LiOH reacted = (2/1)* moles of Li2CO3
= (2/1)*0.4395
= 0.879 mol

Answer: 0.879 mol