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 much `Cl_(2)` is reqired to prepare 122.5 g of `NaClO_(4)` by above sequencial reactions ?

Options

A. 284 g
B. 213 g
C. 142 g
D. 71 g

Answer

To determine how much Cl₂ is required to prepare 122.5 g of NaClO₄ through the given sequential reactions, we will follow these steps:

### Step 1: Determine the molar mass of NaClO₄
The molar mass of NaClO₄ (sodium perchlorate) can be calculated as follows:
- Sodium (Na): 22.99 g/mol
- Chlorine (Cl): 35.45 g/mol
- Oxygen (O): 16.00 g/mol (4 oxygen atoms)

Calculating the total molar mass:
\[ \text{Molar mass of NaClO₄} = 22.99 + 35.45 + (4 \times 16.00) \]
\[ = 22.99 + 35.45 + 64.00 = 122.44 \, \text{g/mol} \]

### Step 2: Calculate the number of moles of NaClO₄ produced
Using the mass of NaClO₄ given in the question (122.5 g), we can calculate the number of moles:
\[ \text{Moles of NaClO₄} = \frac{\text{mass}}{\text{molar mass}} \]
\[ = \frac{122.5 \, \text{g}}{122.44 \, \text{g/mol}} \approx 1.00 \, \text{mol} \]

### Step 3: Determine the stoichiometry of the reactions
From the reactions provided:
1. **Reaction 1**: \( \text{Cl}_2 + 2 \text{NaOH} \rightarrow 2 \text{NaCl} + \text{NaClO} + \text{H}_2O \)
- 1 mole of Cl₂ produces 1 mole of NaClO.

2. **Reaction 2**: \( 3 \text{NaClO} \rightarrow 2 \text{NaCl} + \text{NaClO}_3 \)
- 3 moles of NaClO produce 1 mole of NaClO₃.

3. **Reaction 3**: \( 4 \text{NaClO}_3 \rightarrow 3 \text{NaClO}_4 + \text{NaCl} \)
- 4 moles of NaClO₃ produce 3 moles of NaClO₄.

### Step 4: Calculate the moles of NaClO needed to produce 1 mole of NaClO₄
From the stoichiometry:
- To produce 1 mole of NaClO₄, we need:
- From Reaction 3: \( \frac{4}{3} \) moles of NaClO₃.
- From Reaction 2: \( \frac{4}{3} \times 3 = 4 \) moles of NaClO.

Thus, to produce 1 mole of NaClO₄, we need 4 moles of NaClO.

### Step 5: Calculate the moles of Cl₂ needed
Since 1 mole of Cl₂ produces 1 mole of NaClO:
- To produce 4 moles of NaClO, we need 4 moles of Cl₂.

### Step 6: Calculate the mass of Cl₂ required
Using the molar mass of Cl₂ (70.90 g/mol):
\[ \text{Mass of Cl₂} = \text{moles} \times \text{molar mass} \]
\[ = 4 \, \text{mol} \times 70.90 \, \text{g/mol} = 283.6 \, \text{g} \]

### Conclusion
The amount of Cl₂ required to prepare 122.5 g of NaClO₄ is approximately **284 g**.