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Write the balanced NET ionic equation for the reaction when AICI3 and Na₂S are mixed in aqueous solution. If no reaction occurs, simply write only NR. Be sure to include the proper phases for all species within the reaction.

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Answer:

2AlCl3(aq) + 3Na2S(aq) = Al2S3(aq) + 6NaCl(aq)

Explanation:

First, you have to write down a complete equation as shown below:

AlCl3 + Na2S → Al2S3 + NaCl.

When you at the above equation is not balanced.

To balance the equation, we have to identify the number of products and reactants.

Al: 1a + 0b = 2c + 0d, we have 2 Al

Cl: 3a + 0b = 0c + 1d, we have 3Cl

Na: 0a + 2b = 0c + 1d, we have 2Na

S: 0a + 1b = 3c + 0d, we have 3S

Then we simplify to obtain the lowest whole integer value:

a = 2 (AlCl3)

b = 3 (Na2S)

c = 1 (Al2S3)

d = 6 (NaCl)

Substitute the values to our first equation giving us,

2AlCl3 + 3Na2S = Al2S3 + 6NaCl,

We then include the states. Since the reaction is in aqueous solution, we insert (aq)

2AlCl3(aq) + 3Na2S(aq) = Al2S3(aq) + 6NaCl(aq)

Since there is an equal number of each element in the reactants and products of 2AlCl3(aq) + 3Na2S(aq) = Al2S3(aq) + 6NaCl(aq), the equation is balanced.

Thus our required equation is

2AlCl3(aq) + 3Na2S(aq) = Al2S3(aq) + 6NaCl(aq)

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