if the yieeld for the following reaction is 85% how many grams of Al2S3 should be used to produce 165 g of Al(OH)3?
Al2S3+6H2O-->2Al(OH)3+3H2S
165 g Al(OH)3 = how many mols.
mols = grams/molar mass.
If the yield is only 85%, then you must produce more than that. How much more?
yield = 0.85 = (actual value/theoretical value) = (?mols/x) and solve for x.
Convert mols (x) Al(OH)3 to mols Al2S3. Then g Al2S3 = mols x molar mass.
To solve this problem, we need to use stoichiometry to determine the amount of reactant (Al2S3) needed to produce a given amount of product (Al(OH)3).
Step 1: Write the balanced chemical equation:
Al2S3 + 6H2O -> 2Al(OH)3 + 3H2S
Step 2: Calculate the molar mass of Al2S3 and Al(OH)3:
Molar mass of Al2S3 = (2 x atomic mass of Al) + (3 x atomic mass of S)
Molar mass of Al(OH)3 = (2 x atomic mass of Al) + (6 x atomic mass of H) + (3 x atomic mass of O)
Step 3: Convert the given mass of Al(OH)3 (165 g) to moles:
moles of Al(OH)3 = mass (g) / molar mass (g/mol)
Step 4: Use stoichiometry to find the moles of Al2S3 needed:
From the balanced equation, we know that 2 moles of Al(OH)3 is produced for every 1 mole of Al2S3. Therefore:
moles of Al2S3 = moles of Al(OH)3 / 2
Step 5: Convert moles of Al2S3 to grams:
mass of Al2S3 = moles of Al2S3 x molar mass of Al2S3
Finally, let's follow these steps to find the answer:
Step 1: The balanced chemical equation is already given as:
Al2S3 + 6H2O -> 2Al(OH)3 + 3H2S
Step 2: The molar mass of Al2S3 can be calculated as:
Molar mass of Al2S3 = (2 x atomic mass of Al) + (3 x atomic mass of S)
= (2 x 26.98 g/mol) + (3 x 32.07 g/mol)
= 101.96 g/mol
The molar mass of Al(OH)3 can be calculated as:
Molar mass of Al(OH)3 = (2 x atomic mass of Al) + (6 x atomic mass of H) + (3 x atomic mass of O)
= (2 x 26.98 g/mol) + (6 x 1.01 g/mol) + (3 x 16.00 g/mol)
= 78.00 g/mol
Step 3: Convert the given mass of Al(OH)3 to moles:
moles of Al(OH)3 = mass (g) / molar mass (g/mol)
= 165 g / 78.00 g/mol
= 2.1154 mol
Step 4: Use stoichiometry to find the moles of Al2S3 needed:
From the balanced equation, we know that 2 moles of Al(OH)3 is produced for every 1 mole of Al2S3. Therefore:
moles of Al2S3 = moles of Al(OH)3 / 2
= 2.1154 mol / 2
= 1.0577 mol
Step 5: Convert moles of Al2S3 to grams:
mass of Al2S3 = moles of Al2S3 x molar mass of Al2S3
= 1.0577 mol x 101.96 g/mol
= 108.00 g
Therefore, to produce 165 g of Al(OH)3 with a yield of 85%, approximately 108 grams of Al2S3 should be used.