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Reaction 2 occurs when an excess of 6 M HCI(aq) solution is added to 100. mL of NaOCl(aą) of unknown concentration. If the reaction goes to completion and 0.010 mol of Cl2(g) is produced, then what was the molarity of the NaOCl(aq) solution?
(A) 0.0010 M
(B) 0.010 M
(C) 0.10 M
(D) 1.0 M

Respuesta :

Answer:

Option C 0.10 M

Explanation:

you are not providing the reaction, but luckily we have data to write the equation. We know that we are taking HCl and NaOCl to produce Cl2, the reaction is as follow:

NaOCl + 2HCl ---------> Cl₂ + NaCl + H₂O

From this reaction, we know that we have excess of acid that is added to 100 mL (0.1 L) of NaOCl, and this produces 0.010 moles of Chlorine.

All we have to do here, is apply stechiometry. We already know that the acid is in excess, so the moles produced would be the moles of the limiting reactant, in this case, the NaOCl.

According to the reaction NaOCl and Cl2 has a mole ratio of 1:1, so the moles of Cl2 would be the same moles that NaOCl lost so:

moles Cl2 = moles NaOCl = 0.010 moles

Then the concentration:

M = 0.010 / 0.1

M = 0.10 M

This is the molarity of NaOCl

Eduard22sly

The molarity of the NaOCl solution is 0.10 M

The correct answer to the question is Option C. 0.10 M

We'll begin by calculating the number of mole of NaOCl required to produce 0.010 mole of Cl₂. This can be obtained as follow:

NaOCl + 2HCl —> Cl₂ + NaCl + H₂O

From the balanced equation above,

1 mole of NaOCl reacted to produce 1 mole of Cl₂.

Therefore

0.010 mole of NaOCl will also react to produce 0.010 mole of Cl₂.

  • Finally, we shall determine the molarity of the NaOCl solution.

Mole of NaOCl = 0.01 mole

Volume = 100 mL = 100 / 1000 = 0.1 L

Molarity of NaOCl =?

Molarity = mole / Volume

Molarity of NaOCl = 0.01 / 0.1

Molarity of NaOCl = 0.1 M

Therefore, the molarity of the NaOCl solution is 0.1 M

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