Question

I need help on 4, this is theoretical yeild and writing the equation

Answer 1

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  

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

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.

-----------

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  

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