Question

AlCl3 + Na2CO3 molecular, ionic, and net ionic equations

Answer 1

Explanation:

Molecular equation:

  A molecular equation:

      Molecular equations shows the molecular formula of compounds or moleucles of elements of all reactants and products.

                      AlCl₃$_{aq}$ + Na₂CO₃$_{aq}$ → NaCl$_{aq}$ + Al₂(CO₃)₃$_{s}$

    Balanced equation :   2AlCl₃ + 3Na₂CO₃ → 6NaCl + Al₂(CO₃)₃

  Ionic equation:

   Water soluble reactants and products splits into their constituent ions while non-ionizing or gaseous reactants and products are represented by formulae or molecules.

        2Al³⁺$_{aq}$ + 6Cl⁻ $_{aq}$+ 6Na⁺$_{aq}$ + 2CO₃²⁻ $_{aq}$ →   6Na⁺$_{aq}$ + 6Cl⁻$_{aq}$ + Al₂(CO₃)₃$_{s}$

 Net ionic equation:

  Ions that are non-reactive are called spectator ions, they are cancelled out to give the net ionic equation.

                         2Al³⁺$_{aq}$ + 6Cl⁻$_{aq}$  →  Al₂(CO₃)₃$_{s}$

Learn more:

Ionic equation brainly.com/question/2947744

#learnwithBrainly

Answer 2

Answer:

Molecular equation is nothing but the balanced chemical equation with the physical states of substances shown which we find out using the solubility rules.

Group IA salts are soluble so $Na_2CO_3$ and NaCl are soluble in water hence (aq) is the physical state.

Chlorides are generally soluble except that of Ag, Pb amd Hg so $AlCl_3$ is soluble.

Carbonates are insoluble except that of Group IA and ammonium ion  

hence $Al_2(CO_3 )_3$ is insoluble and the physical state is (s) which indicates  insoluble solid or a precipitate

Insoluble solid precipitate (s), Liquid (l) and gas (g) do not produce ions.

Salts which are soluble in water (aq) produce ions which are shown in the ionic equation  

Net ionic equation do not contain the common ions which are present  on both the sides of the ionic equation and are called as spectator ions  

Balancing is making the number of atoms of each element same on both the sides (reactant and product side).

To find the number of atoms of each element we multiply coefficient and the subscript  

For example $5Ca_1Cl_2$ contains  

5 ×1 = 5, Ca atoms and

5 × 2 =10, Cl atoms  

If there is a bracket in the chemical formula  

For example $3Ca_3(P_1 O_4 )_2$ we multiply coefficient  × subscript × number outside the bracket to find the number of atoms  (Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )

So

3 × 3 = 9 , Ca atoms  

3 × 1 × 2 = 6, P atoms  

3 × 4 × 2 = 24, O atoms are present.

Answer:

Molecular equation:

$2AlCl_3 (aq)+3Na_2 CO_3 (aq)>Al_2(CO_3 )_3(s)+6NaCl(aq)$

Ionic Equation :

$2Al^{3+}(aq)+6Cl^{1-}(aq)+6Na^{1+}(aq)+3CO_3^{2-}(aq)>1Al_2(CO_3 )_3 (s)+6Na^{1+}(aq)+6Cl^{1-}(aq)$

Net ionic equation :

$2Al^{3+}(aq)+3CO_3^{2-}(aq)>1Al_2(CO_3 )_3 (s)$

Please note: $Cl^-$ and $Na^+$ are spectator ions