Answered

Reaction 2 occurs when an excess of 6 M HCl (aq)solution is added to 100 mL of NaOCl (aq) 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 solution?

(A)0.0010 M

(B)0.010 M

(C)0.10 M

(D)1.0 M

Answer :

anfabba15

Answer:

[NaOCl] = 0.1M   (option c.)

Explanation:

We determine the reaction:

NaOCl(aq) + HCl (aq) →NaOH (aq) +  Cl₂(g)

Ratio is 1:1, so if I produce 0.01 moles of chlorine I used 0.01 mol  of hypochlorite.

Our volume of solution by water (solvent) and NaOCl (solute) was 100 mL, so in order to determine molarity we divide moles by volume (L)

100 mL . 1L/1000mL = 0.1L

Molarity → 0.01 mol /0.1L = 0.1M

The molarity of the NaOCl solution is 0.1 M. The correct answer is (C)0.10 M

Stoichiometry

From the question, we are to determine the molarity of the NaOCl

From balanced chemical equation for reaction 2

NaOCI(aq) + 2 HCl(aq) – Cl2(g) + NaCI(aq) + H20(l)

This means,

1 mole of NaOCl reacts with 2 moles of HCl to give 1 mole of Cl₂ gas

Since,

0.010 mole of Cl₂ gas is produced

Then,

The number of moles of NaOCl that reacted is 0.010 mol

Now, for the molarity of the NaOCl

Using the formula,

[tex]Molarity = \frac{Number\ of\ moles }{Volume}[/tex]

From the given information,

Volume = 100 mL = 0.1 L

Therefore,

[tex]Molarity = \frac{0.010}{0.1}[/tex]

Molarity = 0.1 M

Hence, the molarity of the NaOCl solution is 0.1 M. The correct answer is (C)0.10 M

Learn more on Stoichiometry here: https://brainly.com/question/19920764

Here is the remaining part of the question:

Reaction 1: CaC₂(s) + 2H₂O(l) → C₂H₂(g) + Ca(OH)₂(s)

Reaction 2: NaOCI(aq) + 2HCl(aq) → Cl₂(g) + NaCI(aq) + H₂O(l)

Reaction 3: C₂H₂(g) + Cl₂(g) → C₂H₂Cl₂(g)

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