Precipitation reactions and ionic equations
(look at attachment)

Net ionic equation: 2OH⁻(aq) + Fe²⁺(aq)  → Fe(OH)₂(s)

Net ionic equation: 2K⁺(aq) + CH₃COO⁻ (aq) + 2Na⁺(aq) + S⁻²(aq)  → NO REACTION

Net ionic equation: CO₃⁻²(aq) + Pb⁺²(aq) → PbCO₃(s)

Explanation:

a. Solutions of calcium hydroxide and Iron (II) chloride are mixed:

We identify the reactants:

Ca(OH)₂ , FeCl₂

In excess, the Fe(OH)₂ can make precipitate

Salts from chlorides with elements from group II are soluble.

The reaction is: Ca(OH)₂(aq) + FeCl₂(aq) →  Fe(OH)₂(s)  +  CaCl₂(aq)

Ca(OH)₂(aq) + FeCl₂(aq) →  Fe(OH)₂(s)  +  CaCl₂(aq)

We dissociate the compounds, except for the solid

Ca²⁺(aq) + 2OH⁻(aq) + Fe²⁺(aq) + 2Cl⁻(aq)  →  Fe(OH)₂(s) + Ca²⁺ + 2Cl⁻(aq)

Net ionic equation: 2OH⁻(aq) + Fe²⁺(aq) → Fe(OH)₂(s)

b. Solutions of potassium acetate and sodium sulfide are mixed:

The reactants are: KCH₃COOH and Na₂S

In this case there are no precipitates, because all the salts are soluble

We make the complete reaction:

2KCH₃COO (aq) + Na₂S(aq)  → K₂S(aq) + 2NaCH₃COO (aq)

Net ionic equation is:

2K⁺(aq) + CH₃COO⁻ (aq) + 2Na⁺(aq) + S⁻²(aq)  → 2K⁺(aq) + S⁻²(aq) + 2Na⁺(aq) + CH₃COO⁻ (aq)

c. Solutions of ammonium carbonate and lead(II) nitrate are mixed:

In this case, the reactants are: (NH₄)₂CO₃ and Pb(NO₃)₂

All salts from nitrate are soluble.

Carbonate makes a precipitate when it bonds Pb.

The complete reaction is:

(NH₄)₂CO₃(aq) + Pb(NO₃)₂(aq) → PbCO₃(s) + 2NH₄NO₃(aq)

We dissociate all of the compounds, except for the solid in order to make the net ionic equation:

2NH₄⁺(aq) +CO₃⁻²(aq) + Pb⁺²(aq) +2NO₃⁻(aq) → PbCO₃(s) + 2NH₄⁺ (aq) + 2NO₃⁻(aq)

The net ionic equation is: CO₃⁻²(aq) + Pb⁺²(aq) → PbCO₃(s)

The ions that are repeated, are called spectators ions. We all cancel them.

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

its first and third is your answer i did the qustion the fourth isnt right