AlCl3 and Na2SO4

Balanced Equation

Total Ionic Equation

Net Ionic Equation

a) Balanced Equation

Al₂ (SO₄) ₃ + 6NaCl → 2AlCl₃ + 3Na₂SO₄

b) Total Ionic Equation

Al₂²⁺ + 3SO₄²⁻ + 6Na⁺ + 6Cl⁻ → 2Al²⁺ + 6Cl⁻ + 6Na⁺ + SO₄²⁻

c) Net Ionic Equation

All the ions in the above equation are spectator, therefore no net equation.

2AlCl3 (aq) + 3Na2SO4 (aq) → Al2 (SO4) 3 (aq) + 6NaCl (aq)

2Al^3 + (aq) + 3Cl - (aq) + 6Na + (aq) + 3SO4^2 - (aq) → 2Al^3 + (aq) + 3SO4^2 - (aq)

+ 6NA + (aq) + 6Cl - (aq)

we do not have a net ionic equation

Explanation:

Step 1: Data given

The reactants are : AlCl3 and Na2SO4

Step 2: The unbalanced equation

AlCl3 (aq) + Na2SO4 (aq) → Al2 (SO4) 3 (aq) + NaCl (aq)

Step 3: Balancing the equation

AlCl3 (aq) + Na2SO4 (aq) → Al2 (SO4) 3 (aq) + NaCl (aq)

On the left side we have 1x Al (in AlCl3), on the right side we have 2x Al (in Al2 (SO4) 3). To balance the amount of Al on both sides, we have to multiply AlCl3 by 2.

2AlCl3 (aq) + Na2SO4 (aq) → Al2 (SO4) 3 (aq) + NaCl (aq)

On the left side we have 1x SO4, on the right side we have 3x SO4

To balance the amount of SO4 on both sides we have to multiply Na2SO4 on the left side by 3

2AlCl3 (aq) + 3Na2SO4 (aq) → Al2 (SO4) 3 (aq + NaCl (aq)

On the left side we have 6x Na, on the right side we have 1x Na

To balance the amount of Na on both sides we have to multiply NaCl on the right side by 6

2AlCl3 (aq) + 3Na2SO4 (aq) → Al2 (SO4) 3 (aq) + 6NaCl (aq)

The total ionic equation

The complete ionic equation is used to describe the chemical reaction while also clearly indicating which of the reactants and/or products exist primarily as ions in aqueous solution.

2Al^3 + (aq) + 3Cl - (aq) + 6Na + (aq) + 3SO4^2 - (aq) → 2Al^3 + (aq) + 3SO4^2 - (aq)

+ 6NA + (aq) + 6Cl - (aq)

The net ionic equation

In a net ionic equation, all spectator ions are completely removed.

For this equation we do not have a net ionic equation