Ammonium perchlorate, `NH_4ClO_4` , used in the solid fuel in the booster rockets on the space shuttle, is prepared from sodium perchlorate, `NaClO_4` , which is produced commercially by the electrolysis of a hot, stirred solution of sodium chloride. How many faradays are required to produce 1.0kg of sodium perchlorate?
`NaCl+4H_2OtoNaClO_4+4H_2`

Options

A. 40.3
B. 18.3
C. 31.6
D. 65.3

Answer

To determine how many Faradays are required to produce 1.0 kg of sodium perchlorate (NaClO₄) from sodium chloride (NaCl) through electrolysis, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Reaction**:
The electrolysis reaction is given as:
\[ \text{NaCl} + 4 \text{H}_2\text{O} \rightarrow \text{NaClO}_4 + 4 \text{H}_2 \]

2. **Determine Oxidation States**:
- In NaCl, sodium (Na) is in the +1 oxidation state and chlorine (Cl) is in the -1 oxidation state.
- In NaClO₄, sodium (Na) remains +1, chlorine (Cl) is in the +7 oxidation state, and oxygen (O) is in the -2 oxidation state.

3. **Calculate Change in Oxidation State**:
The change in oxidation state for chlorine is from -1 to +7, which is a change of:
\[ +7 - (-1) = 8 \]
This means that 8 electrons are required to convert one mole of NaCl to one mole of NaClO₄.

4. **Determine Moles of NaClO₄ in 1 kg**:
The molar mass of NaClO₄ is calculated as follows:
- Na: 22.99 g/mol
- Cl: 35.45 g/mol
- O (4 atoms): 4 × 16.00 g/mol = 64.00 g/mol
\[ \text{Molar mass of NaClO}_4 = 22.99 + 35.45 + 64.00 = 122.44 \text{ g/mol} \]
Now, to find the number of moles in 1 kg (1000 g):
\[ \text{Moles of NaClO}_4 = \frac{1000 \text{ g}}{122.44 \text{ g/mol}} \approx 8.16 \text{ moles} \]

5. **Calculate Faradays Required**:
Since it takes 8 Faradays to produce 1 mole of NaClO₄, the total Faradays required for 8.16 moles is:
\[ \text{Faradays required} = 8 \text{ Faradays/mole} \times 8.16 \text{ moles} \approx 65.28 \text{ Faradays} \]

6. **Final Answer**:
Rounding to one decimal place, the total Faradays required to produce 1.0 kg of sodium perchlorate is approximately:
\[ \boxed{65.3 \text{ Faradays}} \]