How much `Cl_(2)` in needed to prepare `106.5g`of `NaClO_(3)` by the above sequence?

Options

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

Explanation

To determine how much chlorine gas (`Cl2`) is needed to prepare `106.5 g` of sodium chlorate (`NaClO3`), we will follow these steps:

### Step 1: Write down the relevant chemical equations
The reactions involved in the preparation of `NaClO3` from `Cl2` are:
1. \( \text{Cl}_2 + 2 \text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O} \) (Equation 1)
2. \( 3 \text{NaClO} \rightarrow \text{NaCl} + \text{NaClO}_3 \) (Equation 2)
3. \( 4 \text{NaClO}_3 \rightarrow 3 \text{NaClO}_4 + \text{NaCl} \) (Equation 3)

### Step 2: Determine the molar mass of `NaClO3`
To find the number of moles of `NaClO3`, we first calculate its molar mass:
- Sodium (Na) = 23 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Oxygen (O) = 16 g/mol

Calculating the molar mass:
\[ \text{Molar mass of NaClO}_3 = 23 + 35.5 + (16 \times 3) = 23 + 35.5 + 48 = 106.5 \text{ g/mol} \]

### Step 3: Calculate the number of moles of `NaClO3`
Using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
Substituting the values:
\[ \text{Number of moles of NaClO}_3 = \frac{106.5 \text{ g}}{106.5 \text{ g/mol}} = 1 \text{ mole} \]

### Step 4: Relate moles of `Cl2` to moles of `NaClO3`
From the reactions:
- From Equation 1, 1 mole of `Cl2` produces 1 mole of `NaClO` (and thus can be used to produce `NaClO3`).
- From Equation 2, 3 moles of `NaClO` produce 1 mole of `NaClO3`.

Thus, to produce 1 mole of `NaClO3`, we need:
\[ \text{Moles of } Cl_2 = 3 \text{ moles of } Cl_2 \text{ (since 1 mole of } Cl_2 \text{ gives 1 mole of } NaClO \text{ and 3 moles of } NaClO \text{ gives 1 mole of } NaClO_3) \]

### Step 5: Calculate the mass of `Cl2`
The molar mass of `Cl2` is:
\[ \text{Molar mass of } Cl_2 = 2 \times 35.5 = 71 \text{ g/mol} \]
Now, we calculate the mass of `Cl2` needed for 3 moles:
\[ \text{Mass of } Cl_2 = \text{Number of moles} \times \text{Molar mass} = 3 \text{ moles} \times 71 \text{ g/mol} = 213 \text{ g} \]

### Final Answer
To prepare `106.5 g` of `NaClO3`, `213 g` of `Cl2` is needed.

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