Consider the following series of reactions :
`Cl_(2)+2NaOH to NaCl+NaClO+H_(2)O`
`3NaClO to 2NaCl+NaClO_(3)`
`4NaClO_(3) to 3NaClO_(4)+NaCl`
How many moles of NaCl will be formed by using 1 mole `Cl_(2)` and other reagents in excess ?

Options

A. `(1)/(12)` mole
B. 1.67 mole
C. 1.75 mole
D. 0.75 mole

Answer

To determine how many moles of NaCl will be formed from 1 mole of Cl₂ and excess reagents, we can analyze the given reactions step by step.

### Step 1: Analyze the first reaction
The first reaction is:
\[ \text{Cl}_2 + 2 \text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O} \]

From this reaction, we see that:
- 1 mole of Cl₂ produces 1 mole of NaCl.
- Therefore, if we start with 1 mole of Cl₂, we will produce 1 mole of NaCl.

### Step 2: Analyze the second reaction
The second reaction is:
\[ 3 \text{NaClO} \rightarrow 2 \text{NaCl} + \text{NaClO}_3 \]

From the first reaction, we produced 1 mole of NaClO. Now we need to find out how much NaCl can be produced from this 1 mole of NaClO.

Using the stoichiometry of the second reaction:
- From 3 moles of NaClO, we get 2 moles of NaCl.
- Therefore, from 1 mole of NaClO, we can calculate the moles of NaCl produced:
\[ \text{Moles of NaCl from NaClO} = \frac{2}{3} \text{ moles of NaCl} \]

### Step 3: Analyze the third reaction
The third reaction is:
\[ 4 \text{NaClO}_3 \rightarrow 3 \text{NaClO}_4 + \text{NaCl} \]

From the second reaction, we produced 1 mole of NaClO₃. Now we need to find out how much NaCl can be produced from this 1 mole of NaClO₃.

Using the stoichiometry of the third reaction:
- From 4 moles of NaClO₃, we get 1 mole of NaCl.
- Therefore, from 1 mole of NaClO₃, we can calculate the moles of NaCl produced:
\[ \text{Moles of NaCl from NaClO}_3 = \frac{1}{4} \text{ moles of NaCl} \]

### Step 4: Calculate the total moles of NaCl
Now we can sum up all the moles of NaCl produced from each step:
1. From the first reaction: 1 mole of NaCl
2. From the second reaction: \(\frac{2}{3}\) moles of NaCl
3. From the third reaction: \(\frac{1}{4}\) moles of NaCl

Now, we need to find a common denominator to add these fractions. The least common multiple of 1, 3, and 4 is 12.

- Convert each term to have a denominator of 12:
- \(1 = \frac{12}{12}\)
- \(\frac{2}{3} = \frac{8}{12}\)
- \(\frac{1}{4} = \frac{3}{12}\)

Now, add them together:
\[ \text{Total NaCl} = \frac{12}{12} + \frac{8}{12} + \frac{3}{12} = \frac{23}{12} \text{ moles of NaCl} \]

### Final Answer
The total number of moles of NaCl formed is \(\frac{23}{12}\) moles.

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