Answer:
2.26 g
Explanation:
Data given:
mass of Sulfuric acid = 3.24 g
mass of Aluminum hydroxide = 0.945 g
Theoretical yield of Aluminum sulfate = ?
Solution:
First we look for the balance reaction
Reaction
3H₂SO₄ + 2Al(OH)₃ ----------> Al₂(SO₄)₃ + 6H₂O
Now we look for the limiting reactant on which the amount of aluminum sulfate depends
So,
3H₂SO₄ + 2Al(OH)₃ ----------> Al₂(SO₄)₃ + 6H₂O
3 mol 2 mol
from above reaction it is clear that 3 mole of H₂SO₄ combine with 2 mole of Al(OH)₃
Convert moles to mass
molar mass of Al(OH)₃ = 27 + 3(16 + 1)
molar mass of Al(OH)₃ = 27 + 45
molar mass of Al(OH)₃ =72 g/mol
molar mass of H₂SO₄ = 2(1) + 32 + 4(16)
molar mass of H₂SO₄ = 2 + 32 + 64
molar mass of H₂SO₄ = 98 g/mol
So,
3H₂SO₄ + 2Al(OH)₃ ----------> Al₂(SO₄)₃ + 6H₂O
3 mol (98 g/mol) 2 mol (72 g/mol)
294 g 144 g
So its clear from the reaction that 294 g of H₂SO₄ react with 144 g of Al(OH)₃
now if we look at the given amounts the amount Al(OH)₃ is less then the amount of H₂SO₄
So, for Al(OH)₃ if we calculate the needed amount of H₂SO₄
So apply unity formula
294 g H₂SO₄ ≅ 144 g of Al(OH)₃
X g H₂SO₄ ≅ 0.945 of Al(OH)₃
Do cross multiplication
X g H₂SO₄ = 294 g x 0.945 g / 144 g
X g of H₂SO₄ ≅ 1.93 g
So, 1.93 g of H₂SO₄ will react out of 3.24 grams, the remaining amount of it will be in excess.
So,
Al(OH)₃ will be consumed completely an it will be limiting reactant.
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Now to Calculate for the theoretical yield
First we look for the balance reaction
Reaction
3H₂SO₄ + 2Al(OH)₃ ----------> Al₂(SO₄)₃ + 6H₂O
Now we look for the mole mole ration of Al(OH)₃ to the amount of aluminum sulfate produced
So,
3H₂SO₄ + 2Al(OH)₃ ----------> Al₂(SO₄)₃ + 6H₂O
2 mol 1 mole
from above reaction it is clear that 1 mole of Al₂(SO₄)₃ produce by 2 mole of Al(OH)₃
As we know that
2 mole of Al(OH)₃ = 144 g
So,
if 144 g of Al(OH)₃ gives 1 mole Al₂(SO₄)₃ then how many moles of Al₂(SO₄)₃ will be produces by 0.945 g Al(OH)₃
So apply unity formula
144 g of Al(OH)₃ ≅ 1 mole of Al₂(SO₄)₃
0.945 g of Al(OH)₃ ≅ X mole of Al₂(SO₄)₃
Do cross multiplication
X mole of Al₂(SO₄)₃ = 0.945 g x 1 mole / 144 g
X mole of Al₂(SO₄)₃ = 0.0066 moles
So,
0.945 g of Al(OH)₃ produce 0.0066 mole of Al₂(SO₄)₃
Now conver moles of Al₂(SO₄)₃ to mass
Formula used:
mass in grams = no. of moles x molar mass . . . . . . (1)
molar mass of Al₂(SO₄)₃ = 2(27) + 3(32 +4(16))
molar mass of Al₂(SO₄)₃ = 54 + 3 (32 +64)
molar mass of Al₂(SO₄)₃ = 54 + 3 (96)
molar mass of Al₂(SO₄)₃ = 54 + 288
molar mass of Al₂(SO₄)₃ =342 g/mol
Put values in equation 1
mass in grams = 0.0066 g x 342 g/mol
mass in grams = 2.26 g
So the theoretical yield of Aluminum sulfate (Al₂(SO₄)₃ ) is 2.26 g
