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Hint: Barium chloride stands for$BaC{{l}_{2}}$, aluminum sulphate stands for $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$, barium sulphate stands for $BaS{{O}_{4}}$ and last aluminum chloride stands for $AlC{{l}_{_{3}}}$. Arrange them, chemical reactions and balance them.
The symbolic representation of barium is $Ba$.
The symbolic representation of chloride is $Cl$.
The symbolic representation of aluminum is $Al$.
The symbolic representation of sulphate is $S{{O}_{4}}$.
Complete step-by-step answer:
Barium chloride is represented as $BaC{{l}_{2}}$ barium has oxidation state and chlorine has oxidation state. Therefore 1 barium combines with 2 chlorides.
Aluminum sulphate is represented by $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ as aluminum has oxidation state and sulphate has oxidation state. So, 3 aluminum reacts with 3 molecules of sulphate.
When barium chloride reacts with aluminum sulphate it gives aluminum chloride and barium sulphate as a product.
Barium sulphate is represented as $BaS{{O}_{4}}$ barium has oxidation state and sulphate has oxidation state. Therefore 1 barium combines with 1 Sulphur.
Aluminium chloride is represented by $AlC{{l}_{_{3}}}$ as aluminum has oxidation state and chlorine has oxidation state. So, 1 aluminum reacts with 3 chlorines.
So overall reaction is:
$BaC{{l}_{2}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\to AlC{{l}_{3}}+BaS{{O}_{4}}$
Now, balancing the equation as the number of elements as they are not equal on both sides.
So the balanced equation is:
$3BaC{{l}_{2}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\to 2AlC{{l}_{3}}+3BaS{{O}_{4}}$
Now labeling the physical state of each reactant and product:
$3BaC{{l}_{2}}(aq)+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}(aq)\to 2AlC{{l}_{3}}(aq)+3BaS{{O}_{4}}(s)$
It is given that barium sulphate precipitates therefore it is solid and others are aqueous in their physical state.
Note: Chemical equation is written by the chemical name of each compound present as reactant or product arranging them in an equation followed by balancing of equation as stoichiometry should be the same. After balancing the equation, the physical state of each reactant and product is written.
The symbolic representation of barium is $Ba$.
The symbolic representation of chloride is $Cl$.
The symbolic representation of aluminum is $Al$.
The symbolic representation of sulphate is $S{{O}_{4}}$.
Complete step-by-step answer:
Barium chloride is represented as $BaC{{l}_{2}}$ barium has oxidation state and chlorine has oxidation state. Therefore 1 barium combines with 2 chlorides.
Aluminum sulphate is represented by $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ as aluminum has oxidation state and sulphate has oxidation state. So, 3 aluminum reacts with 3 molecules of sulphate.
When barium chloride reacts with aluminum sulphate it gives aluminum chloride and barium sulphate as a product.
Barium sulphate is represented as $BaS{{O}_{4}}$ barium has oxidation state and sulphate has oxidation state. Therefore 1 barium combines with 1 Sulphur.
Aluminium chloride is represented by $AlC{{l}_{_{3}}}$ as aluminum has oxidation state and chlorine has oxidation state. So, 1 aluminum reacts with 3 chlorines.
So overall reaction is:
$BaC{{l}_{2}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\to AlC{{l}_{3}}+BaS{{O}_{4}}$
Now, balancing the equation as the number of elements as they are not equal on both sides.
So the balanced equation is:
$3BaC{{l}_{2}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\to 2AlC{{l}_{3}}+3BaS{{O}_{4}}$
Now labeling the physical state of each reactant and product:
$3BaC{{l}_{2}}(aq)+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}(aq)\to 2AlC{{l}_{3}}(aq)+3BaS{{O}_{4}}(s)$
It is given that barium sulphate precipitates therefore it is solid and others are aqueous in their physical state.
Note: Chemical equation is written by the chemical name of each compound present as reactant or product arranging them in an equation followed by balancing of equation as stoichiometry should be the same. After balancing the equation, the physical state of each reactant and product is written.
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