Answer:
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
