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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.)

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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:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     .....(1)

Given mass of lithium oxide = 250 g

Molar mass of lithium oxide = 30 g/mol

Putting values in equation 1, we get:

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

For the given chemical equation:

[tex]Li_2O(s)+H_2O(g)\rightarrow 2LiOH(s)[/tex]

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 = [tex]\frac{1}{1}\times 8.33=8.33mol[/tex] 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:

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

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

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

Number of moles:

  • It is defined as the ratio of given mass over molar mass.
  • It is given by:

[tex]\text{Number of moles}=\frac{\text {Given mass}}{\text{ Molar mass}}[/tex]

Given:

Mass of lithium oxide = 250 g

Molar mass of lithium oxide = 30 g/mol

  • On substituting the values in the above formula:

[tex]\text{Number of moles}=\frac{\text {Given mass}}{\text{ Molar mass}}\\\\\text{Number of moles}=\frac{250g}{30g/mol}\\\\ \text{Number of moles}=8.33 moles[/tex]

For the given chemical equation:

[tex]Li_2O(s) + H_2O(g) --> 2 LiOH(s)[/tex]

We observe that:

1 mole of lithium oxide reacts with 1 mole of water.

So, 8.33 moles of lithium oxide will react with =[tex]\frac{1}{1}*8.33=8.33[/tex] moles  of water

  • Calculation for mass of water:

Molar mass of water = 18 g/mol

Moles of water = 8.33 moles

[tex]\text{Number of moles}=\frac{\text {Given mass}}{\text{ Molar mass}}\\\\\text{Mass}= \text{Number of moles}*\text{ Molar mass}\\\\\text{Mass}=8.33mol*18g/mol\\\\\text{Mass}=149.94g[/tex]

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

Find more information about Number of moles here:

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