Question

Consider the following unbalanced chemical equation.LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)If 83.1 grams of lithium hydroxide reacts with excess carbon dioxide, what mass of lithium carbonate will be produced?

Answer 1

In this question, we have the following reaction:

2 LiOH + CO2 --> Li2CO3 + H2O

We have:

83.1 grams of LiOH

We want to know how much of Li2CO3 will be produced

In order to find out, first we need to find the number of moles of LiOH present in 83.1 grams, we will use its molar mass, 23.95g/mol to find it

23.95 grams = 1 mol

83.1 grams = x moles

23.95x = 83.1

x = 83.1/23.95

x = 3.5 moles of LiOH in 83.1 grams

Now that we have the number of moles of LiOH, we will use the molar ratio between the two compounds to find the number of moles of Li2CO3, the molar ratio is 2:1

3.5/2 = 1.75 moles of Li2CO3

Now we use the molar mass of Li2CO3, 73.89g/mol, to find the final mass:

73.89 grams = 1 mol

x grams = 1.75 moles

x = 1.75 * 73.89

x = 129.3 grams

The final mass will be 129.3 grams of Li2CO3