How many grams of Li2O must be carried on board to remove 85.0kg of water?
1.41×105 g
To determine how many grams of Li2O are needed to remove 85.0kg of water, we need to calculate the molar mass of Li2O and then convert the mass of water into moles.
Step 1: Calculate the molar mass of Li2O.
The molar mass of Li2O can be calculated by adding the atomic masses of its constituent elements.
- Atomic mass of Li = 6.941 g/mol
- Atomic mass of O = 16.00 g/mol
Molar mass of Li2O = (2 * atomic mass of Li) + atomic mass of O
= (2 * 6.941 g/mol) + 16.00 g/mol
= 13.882 g/mol + 16.00 g/mol
= 29.882 g/mol
Step 2: Convert the mass of water into moles.
Using the molar mass of water (H2O), we can convert its mass into moles.
- Molar mass of H2O = 18.015 g/mol (approximately 18.02 g/mol since hydrogen has a molar mass of around 1 g/mol).
Moles of water = Mass of water / Molar mass of water
= 85.0 kg / (18.02 g/mol)
= 4,717.32 mol (approximately)
Step 3: Calculate the amount of Li2O needed to remove the water.
Since we have a 1:1 stoichiometric ratio between Li2O and H2O (i.e., 1 mol of Li2O reacts with 1 mol of H2O), the number of moles of Li2O needed would be equal to the number of moles of H2O.
Moles of Li2O needed = Moles of water
= 4,717.32 mol
Step 4: Convert the moles of Li2O into grams.
Using the molar mass of Li2O calculated earlier, we can convert moles into grams.
Mass of Li2O = Moles of Li2O needed * Molar mass of Li2O
= 4,717.32 mol * 29.882 g/mol
= 140,919.98 g (approximately)
Therefore, approximately 140,920 grams of Li2O must be carried on board to remove 85.0kg of water.
To determine the number of grams of Li2O required to remove 85.0kg of water, we need to use the molar mass of Li2O and stoichiometry.
First, let's calculate the number of moles of water. We know the mass of water, which is 85.0kg. To convert this mass to moles, we need to divide by the molar mass of water, which is approximately 18.015 g/mol.
Number of moles of water = Mass of water / Molar mass of water
= 85,000g / 18.015 g/mol
≈ 4719.3 mol
Next, we need to determine the stoichiometry of the reaction between Li2O and water. The balanced equation for the reaction is:
Li2O + H2O -> 2LiOH
From the equation, we can see that 1 mole of Li2O reacts with 1 mole of water to produce 2 moles of LiOH.
Using the stoichiometry, we can calculate the number of moles of Li2O required to react with the moles of water:
Number of moles of Li2O = (Number of moles of water) / 1
≈ 4719.3 mol
Finally, we need to convert the moles of Li2O to grams by multiplying by the molar mass of Li2O, which is approximately 29.881 g/mol.
Number of grams of Li2O = (Number of moles of Li2O) * (Molar mass of Li2O)
≈ 4719.3 mol * 29.881 g/mol
≈ 140,880 g
Therefore, approximately 140,880 grams of Li2O must be carried onboard to remove 85.0kg of water.
Li2O + HOH ==> 2LiOH
mols H2O = 85,000/18 = estimated 4722
Now convert mols H2O to mols Li2O using the coefficients in the balanced equation. That's estimated 4722 x (1 mol Li2O/1 mol H2O) = about 4722 mols Li2O.
Now convert that to g Li2O. g = mols x molar mass.
Go back and confirm my estimates.