Use the balanced equation for the reaction of lithium oxide with water:
Li2O(s) + H2O(g) --> 2LiOH(s)
to determine how many grams of H2O can be removed from the air by 250g of Li2O. Show your step-by-step solution with unit conversions.
Step 1: Convert 250g of Li2O to moles
250g Li2O x (1 mol Li2O/22.99g Li2O) = 10.9 moles Li2O
Step 2: Use the mole ratio from the balanced equation to determine the moles of H2O
10.9 moles Li2O x (1 mol H2O/2 mol Li2O) = 5.45 moles H2O
Step 3: Convert moles of H2O to grams
5.45 moles H2O x (18.02g H2O/1 mol H2O) = 97.7g H2O
Alright, let's calculate how many grams of H2O can be removed from the air by 250g of Li2O.
First, we need to determine the molar mass of Li2O. Lithium (Li) has a molar mass of approximately 6.94 g/mol, and oxygen (O) has a molar mass of about 16.00 g/mol. Therefore, the molar mass of Li2O is:
2(6.94 g/mol) + 16.00 g/mol = 29.88 g/mol.
Next, we can calculate the number of moles of Li2O using its molar mass. To do this, we divide the given mass by the molar mass:
250 g / 29.88 g/mol = 8.36 mol.
According to the balanced equation, 1 mole of Li2O reacts with 1 mole of H2O. Therefore, the number of moles of H2O that can be removed from the air is also 8.36 mol.
Finally, we can calculate the mass of H2O using its molar mass. The molar mass of H2O is around 18.02 g/mol. Thus:
8.36 mol x 18.02 g/mol ≈ 150.35 g.
So, approximately 150.35 grams of H2O can be removed from the air by 250 grams of Li2O.
To determine how many grams of H2O can be removed from the air by 250g of Li2O, we need to use stoichiometry based on the balanced equation:
Li2O(s) + H2O(g) -> 2LiOH(s)
Step 1: Calculate the molar mass of Li2O and H2O.
The molar mass of Li2O = 6.94 (Li) x 2 + 16.00 (O) = 30.94 g/mol
The molar mass of H2O = 1.01 (H) x 2 + 16.00 (O) = 18.02 g/mol
Step 2: Determine the number of moles of Li2O.
Number of moles = mass / molar mass
Number of moles of Li2O = 250 g / 30.94 g/mol = 8.08 mol
Step 3: Use the stoichiometry to determine the moles of H2O.
From the balanced equation, we can see that 1 mol of Li2O reacts with 1 mol of H2O.
So, moles of H2O = moles of Li2O
Step 4: Convert moles of H2O to grams.
Grams of H2O = moles of H2O x molar mass of H2O
Grams of H2O = 8.08 mol x 18.02 g/mol = 145.81 g
Therefore, 250 grams of Li2O can remove 145.81 grams of H2O from the air.
To determine how many grams of H2O can be removed from the air by 250g of Li2O, we need to use stoichiometry. Stoichiometry is a method used to relate the quantities of substances in a chemical reaction.
Step 1: Determine the molar mass of Li2O.
Li has a molar mass of approximately 6.94 g/mol, and O has a molar mass of approximately 16.00 g/mol. Therefore, the molar mass of Li2O is:
2(6.94 g/mol) + 16.00 g/mol = 29.88 g/mol (approximately 29.9 g/mol)
Step 2: Calculate the number of moles of Li2O.
Using the molar mass calculated in step 1, we can now calculate the number of moles of Li2O in 250g:
Number of moles = Mass / Molar mass
Number of moles of Li2O = 250g / 29.9 g/mol = 8.36 mol (approximately 8.4 mol)
Step 3: Use the stoichiometry ratio from the balanced chemical equation.
From the balanced equation, we see that 1 mol of Li2O reacts with 1 mol of H2O to produce 2 mol of LiOH.
Step 4: Calculate the number of moles of H2O that can be removed from the air.
Since the stoichiometric ratio is 1:1 between Li2O and H2O, the number of moles of H2O that can be removed is also 8.4 mol.
Step 5: Convert moles of H2O to grams.
To convert moles of H2O to grams, we need to multiply by the molar mass of H2O. The molar mass of H2O is approximately 18.02 g/mol.
Mass of H2O = Number of moles of H2O * Molar mass of H2O
Mass of H2O = 8.4 mol * 18.02 g/mol = 151.68 g (approximately 151.7 g)
Therefore, approximately 151.7 grams of H2O can be removed from the air by 250g of Li2O.