Question

Lithium oxide (Li2O) can be used to remove water from air. The balanced equation for the reaction of lithium oxide with water is: Li2O(s) + H2O(g) --> 2 LiOH(s) List the steps that you would have to follow to determine how many grams of water can be removed from the air by 250 g of Li2O, and identify the links between quantities. (Example: If you wanted to convert from μmol of a substance to grams, the steps would be μmol --> mol --> grams. The link between μmol and mol would be 1 μmol = 1E–6 mol, and the link between mol and grams would be the molar mass.)

Answer 1

Answer: The mass of water that can be removed from air by given amount of lithium oxide is 149.94 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

Given mass of lithium oxide = 250 g

Molar mass of lithium oxide = 30 g/mol

Putting values in equation 1, we get:

$\text{Moles of lithium oxide}=\frac{250g}{30g/mol}=8.33mol$

For the given chemical equation:

$Li_2O(s)+H_2O(g)\rightarrow 2LiOH(s)$

By Stoichiometry of the reaction:

1 mole of lithium oxide reacts with 1 mole of water

So, 8.33 moles of lithium oxide will react with = $\frac{1}{1}\times 8.33=8.33mol$ of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 8.33 moles

Putting values in equation 1, we get:

$8.33mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(8.33mol\times 18g/mol)=149.94g$

Hence, the mass of water that can be removed from air by given amount of lithium oxide is 149.94 grams