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)

0.947 mol LiOH

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

6.4 cm (max) - 25 cm (min)

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.