Question

How many moles of lithium hydroxide would be required to produce 35.0 g of Li₂CO₃ in the following chemical reaction? 2 LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l)

Answer 1

0.947 mol LiOH

Step-by-step explanation:

2 LiOH + CO₂ ⇒ Li₂CO₃ + H₂O

Convert grams of Li₂CO₃ to moles. The molar mass is 73.891 g/mol.

(35.0 g Li₂CO₃)/(73.891 g/mol Li₂CO₃) = 0.4737 mol Li₂CO₃

Use the mole ratio between LiOH and Li₂CO₃ to covert moles of Li₂CO₃ to moles of LiOH.

0.4737 mol Li₂CO₃ × (2 mol LiOH/1 mol Li₂CO₃) = 0.9474 mol LiOH

Round for proper significant figures.

0.9474 ≈ 0.947 mol LiOH