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Question

Barium chloride reacts with aluminium sulfate to give aluminium chloride and a precipitate of barium sulfate.

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Solution

Aluminum sulfate + Barium chloride → Aluminum chloride + Barium sulfate.

Al2(SO4)3 + BaCl2 ------> AlCl3 + BaSO4

Balancing the equation

Step 1: Calculate the number of atoms on both sides

Atom No. of atoms on the Reactant sideNo. of atoms on the Product side
Al 2 1
S 3 1
O 12 4
Ba 11
Cl 2 3



Step 3: Balancing the number of atoms on both sides


1. Here, The No. of atoms on both sides is not equal.
2. So let's balance the atom which is in larger amounts.
3. oxygen is the element with a large number on the reactants side with12 and 4 on the product side. so make the 4 to 12 by putting 3 as a coefficient before it
4. Hence, by putting 4 as the coefficient before barium sulfate, barium and sulfur become three in number, so sulfur is balanced since it is 3 on the reactant side.
5. Balance the barium atoms as 3 on both sides by putting 3 as the coefficient before barium chloride.
6. Hence, chlorine becomes 6 in number on the reactant side, let's balance it by putting 2 as the coefficient before aluminum chloride.
7. Now both aluminum and chlorine are balanced.

Step 2: Number of atoms on both sides after balancing


Atoms No. of atoms on the Reactant side No. of atoms on the Product side
Al 2 2
S 3 3
O 12 12
Ba 3 3
Cl 6 6

On balancing the equation:

Al2(SO4)3 + 3BaCl2 ------> 2AlCl3 + 3BaSO4


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