Question

How many grams of aluminum sulfate form when 3.90 grams of aluminum is placed in 13.65 grams of sulfuric acid?

Answer 1

Answer:

Mass = 17.12 g

Explanation:

Given data:

Mass of Al = 3.90 g

Mass of H₂SO₄ = 13.65

Mass of aluminium sulfate = ?

Solution:

Chemical equation:

3H₂SO₄ + 2Al   →  Al₂(SO₄)₃ + 3H₂

Now we will calculate the number of moles of each reactant.

Moles of H₂SO₄:

Number of moles = mass/ molar mass

Number of moles = 13.65 g/ 98.079 g/mol

Number of moles = 0.14 mol

Moles of Al:

Number of moles = mass/ molar mass

Number of moles = 3.90 g/ 27 g/mol

Number of moles = 0.14 mol

Now we will compare the moles of aluminium sulfate with sulfuric acid and aluminium.

                           H₂SO₄       :         Al₂(SO₄)₃

                               3            :              1

                           0.14            :          1/3×0.14 = 0.05

                              Al            :           Al₂(SO₄)₃

                               2            :             1

                            0.14           :        1/2×0.14 = 0.07

The number of moles of aluminium sulfate produced by sulfuric acid are less so it will limiting reactant and limit the amount of aluminium sulfate.

Mass of aluminium sulfate:

Mass = number of moles × molar mass

Mass = 0.05 mol × 342.15 g/mol

Mass = 17.12 g

Answer 2

17.1g

Explanation:

Mass of aluminium = 3.9g

Mass of sulfuric acid = 13.65g

Unknown:

Mass of aluminium sulfate = ?

Solution:

To solve this problem, we need to:

   Convert the mass to moles

   Find the limiting reagent

The limiting reagent is the one in short supply in the reaction. It is important to delineate this reagent because it determines the extent of the reaction.

Also, it is important to obtain the balanced reaction equation:

    2Al    +     3H₂SO₄     →      Al₂(SO₄)₃    +   3 H₂

Number of moles = $\frac{mass}{molar mass}$

  molar mass of Al = 27g/mol

                      H₂SO₄ = 2(1) + 32 + 4(16) = 98g/mol

Number of moles of Al = $\frac{3.9}{27}$ = 0.14mole

Number of moles of H₂SO₄ = $\frac{13.65}{98}$ = 0.14mole

   2 mole of Al reacts with 3 mole of sulfuric acid

   0.14 mole of Al will react with $\frac{0.14 x 3}{2}$ =  0.21mole

      but the amount of sulfuric acid given is 0.14 mole this suggest the sulfuric acid is in short supply. It is the limiting reagent.

To find the number of mole of aluminium sulfate;

             3 mole of sulfuric acid will produce 1 mole of the sulfate

             0.14 mole of the sulfuric acid will produce $\frac{0.14}{3}$ = 0.05mole

Mass of the sulfate = number of moles x molar mass

molar mass of  Al₂(SO₄)₃ = 2(27) + 3(32 + 4(16)) = 342g/mole

 Mass = 0.05 x 342 = 17.1g

learn more:

Molar mass brainly.com/question/2861244

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