Question

How many moles of aluminum ions are present in 58.0 mL of a 0.0100 M solution of aluminum sulfate?

The answer is ___________ mol of Al3+(aq).

Answer 1

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Answer 2

Given:

Volume of aluminum sulfate = 58.0ml

Concentration of aluminum sulfate = 0.01M

To find:

Moles of aluminum ion in aluminum sulfate

Solution:

we known,

                $Molarity = \frac{n}{V} \times 1000$

here, V = volume of solution (58ml)

        n = number of moles

so, number of moles will be

     $n = \frac{molarity \times volume}{1000}$

     $n = \frac{0.01 \times 58}{1000}$

    n = 0.00058 moles

So, number of moles of Al₂(SO₄)₃ is 0.00058 moles.

Now,

     2Al⁺² + 3SO₄⁻² →  Al₂(SO₄)₃

According to stoichiometric,

   1 mole of Al₂(SO₄)₃ formed by Al⁺² moles = 2 moles

   0.00058 moles of Al₂(SO₄)₃ formed by Al⁺² moles = 2 × 0.00058

                                                                                       = 0.00116 moles

Therefore, 0.00116 moles of Al⁺³ ions required.