Seawater is a good source of magnesium. Magnesium is dissolved in the seawater as Mg²⁺. SUGGEST a substance that could be used to precipitate magnesium out of the water. Write a BALANCED CHEMICAL EQUATION and IDENTIFY THE REACTION TYPE to support your thinking

SO₄⁻² (aq)  +  Mg²⁺ (aq)  ⇄  MgSO₄ (s) ↓

Precipitation reaction

Explanation:

Mg²⁻ is one of the metals at group 2 in the Perodic Table. This group is called, the alkaline earth metals.

To make precipitate the Mg²⁺ we can use a salt of sulfate, either flouride or phosphate. Those are three salts that can precipitate this metal.

We use sodium as cathion so, first of all we dissociate them:

Na₂SO₄ (aq) →  2Na⁺  (aq)  +  SO₄⁻² (aq)

NaF (aq) →  Na⁺ (aq) +  F⁻ (aq)

Na₃PO₄ (aq) →  3Na⁺ (aq)  +  PO₄⁻³ (aq)

In all cases, the reaction is of the precipitation type and is conditioned by the Kps, the solubility product constant. Let's bond the anions with Mg²⁺.

SO₄⁻² (aq)  +  Mg²⁺ (aq)  ⇄  MgSO₄ (s) ↓    Kps

2F⁻ (aq)  +  Mg²⁺ (aq)  ⇄   MgF₂ (s) ↓      Kps

2PO₄⁻³ (aq)  +  3Mg²⁺ (aq)  ⇄  Mg₃PO₄ (s) ↓   Kps  

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g