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3 years ago

Sodium (Na) and iodine gas (I2) react to form sodium iodine (NaI) .

Chemistry

2 answers:

Nikolay [14]3 years ago

7 0

Answer:

2Na + I₂ → 2NaI

Explanation:

The sodium and iodine gas react to form the sodium iodine, correct chemical equation is given below:

2Na + I₂ → 2NaI

All other equations are incorrect because,

NaI → 2Na + I₂

In this equation sodium iodine split to gives sodium and iodine.

Na → NaI + I₂

This equation is incorrect because just one element sodium produces sodium iodine and iodine which is impossible.

I₂ → 2NaI + 2Na

This equation is also incorrect because just one element iodine produces sodium iodine and iodine which is impossible, because if there is no sodium atom present in reactant side then it would not appear in product.

7nadin3 [17]3 years ago

6 0

Answer: is 2Na + I2 ---> 2Nal

Explanation: I took the test and the other person is wrong trust me

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Answer:

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3 years ago

Write a net ionic equation for the overall reaction that occurs when aqueous solutions of sodium hydroxide and hydrosulfuric aci

Helga [31]

Answer:

2OH-(aq) + 2H+(aq) → 2H2O(l)

Explanation:

Step 1: Data given

sodium hydroxide = NaOH

hydrosulfuric acid = H2S

Step 2: The unbalanced equation

NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

Step 3: Balancing the equation

On the left side we have 1x Na (in NaOH), on the right side we have 2x Na (in Na2S). To balance the amount of Na on both sides, we have to multiply NaOH on the left side by 2.

2NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

On the left side we have 4x H (2x in NaOH and 2x in H2S), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 2. Now the equation is balanced.

2NaOH(aq) + H2S(aq) → Na2S(aq) + 2H2O(l)

Step 4: The net ionic equation

2Na+(aq) + 2OH-(aq) + 2H+(aq) + S^2-(aq) → 2Na+(aq) + S^2-(aq) + 2H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2OH-(aq) + 2H+(aq) → 2H2O(l)

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