How many moles of lithium hydroxide would be required to produce 38.5 g of Li₂CO₃ in the following chemical reaction? 2 LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l)

Answer 1

In this problem we will be going from grams of product, to moles of product, and finally to moles of reactant.

First we should should start with what we have (grams of Li2CO3) and use that to determine how many moles of the product formed.

Using the grams of Li2CO3 we can figure out how many moles of the product we have. To do this we must use the molar mass of Li2CO3. The molar mass is 73.88 g/mol which I got by adding the molar mass of each element in the chemical equation Li2CO3. The calculation goes as follows:

38.5 Li₂CO₃ x (1 mol Li₂CO₃/ 73.88 g Li₂CO₃) = 0.521 mols of Li₂CO₃

We can use the molar ratios in the chemical formula to determine molar amounts of the different compounds required to complete the chemical reaction.

The compound of interest is LiOH. The integer in front of LiOH is 2, and the integer in front of Li₂CO₃ is 1.The calculation goes as follows:

0.521 mols Li₂CO₃ x (2 mols LiOH / 1 mol Li₂CO₃) = 1.04 mols of LiOH.

Therefore, since the question is asking for mols of LiOH you have reached your answer at 1.04 mols of LiOH.