Consider the following series of reaction:
`Cl_(2) + 2NaOH rarr NaCl + NaClO + H_(2)O`
`3NaClO rarr 2NaCl + NaClO_(3)`.
`4 NaClO_(3) rarr 3NaClO_(4) + NaCl`
How much `Cl_(2)` is needed to prepare `122.5 g NaClO_(4)` by above sequence?

Options

A. `284.0 g`
B. `213.0 g`
C. `142.0g`
D. `71.0 g`

Explanation

To determine how much \( Cl_2 \) is needed to prepare \( 122.5 \, g \) of \( NaClO_4 \) through the given series of reactions, we will follow these steps:

### Step 1: Write down the reactions
The reactions provided are:
1. \( Cl_2 + 2NaOH \rightarrow NaCl + NaClO + H_2O \)
2. \( 3NaClO \rightarrow 2NaCl + NaClO_3 \)
3. \( 4NaClO_3 \rightarrow 3NaClO_4 + NaCl \)

### Step 2: Determine the molar mass of \( NaClO_4 \)
The molar mass of \( NaClO_4 \) can be calculated as follows:
- \( Na \): 23 g/mol
- \( Cl \): 35.5 g/mol
- \( O \): 16 g/mol (4 oxygen atoms)

Calculating the total:
\[ Molar \, mass \, of \, NaClO_4 = 23 + 35.5 + (4 \times 16) = 23 + 35.5 + 64 = 122.5 \, g/mol \]

### Step 3: Calculate the number of moles of \( NaClO_4 \)
Using the mass given:
\[ \text{Number of moles of } NaClO_4 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{122.5 \, g}{122.5 \, g/mol} = 1 \, mol \]

### Step 4: Relate moles of \( NaClO_4 \) to moles of \( Cl_2 \)
From the reactions:
- From reaction 3: \( 4 \, moles \, of \, NaClO_3 \) produce \( 3 \, moles \, of \, NaClO_4 \).
- From reaction 2: \( 3 \, moles \, of \, NaClO \) produce \( 1 \, mole \, of \, NaClO_3 \).
- From reaction 1: \( 1 \, mole \, of \, Cl_2 \) produces \( 1 \, mole \, of \, NaClO \).

Thus, we can derive:
- \( 1 \, mole \, of \, NaClO_4 \) requires \( \frac{4}{3} \, moles \, of \, NaClO_3 \) and \( \frac{4}{3} \times 3 = 4 \, moles \, of \, Cl_2 \).

### Step 5: Calculate the moles of \( Cl_2 \) required
Since we have \( 1 \, mole \, of \, NaClO_4 \):
\[ \text{Moles of } Cl_2 = 4 \times \text{Number of moles of } NaClO_4 = 4 \times 1 = 4 \, moles \]

### Step 6: Calculate the mass of \( Cl_2 \)
The molar mass of \( Cl_2 \) is:
\[ Molar \, mass \, of \, Cl_2 = 2 \times 35.5 = 71 \, g/mol \]
Now, calculating the mass of \( Cl_2 \):
\[ \text{Mass of } Cl_2 = \text{Moles} \times \text{Molar mass} = 4 \, moles \times 71 \, g/mol = 284 \, g \]

### Final Answer
Thus, the amount of \( Cl_2 \) needed to prepare \( 122.5 \, g \) of \( NaClO_4 \) is \( 284 \, g \).

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