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  

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

there is none because earth is stationary and flat.