Given following reaction, `NaClO_3 + Fe rarr O_2 + FeO + NaCl`
In the above reaction 492 L of `O_2` is obtained at 1 atm & 300 K temperature. Determine mass of `NaClO_3` required (in kg). `(R = 0.082 L atm mol^(-1)K^(-1)`

Options

A. 6.13 Kg
B. 4.35 Kg
C. 3.32 Kg
D. 2.13 Kg

Explanation

To determine the mass of NaClO₃ required to produce 492 L of O₂ at 1 atm and 300 K, we can follow these steps:

### Step 1: Calculate the number of moles of O₂ produced using the ideal gas law.

The ideal gas law is given by the formula:
\[ PV = nRT \]

Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.082 L atm K⁻¹ mol⁻¹)
- \( T \) = temperature (in Kelvin)

Given:
- \( P = 1 \) atm
- \( V = 492 \) L
- \( R = 0.082 \) L atm K⁻¹ mol⁻¹
- \( T = 300 \) K

Rearranging the ideal gas law to solve for \( n \):
\[ n = \frac{PV}{RT} \]

Substituting the values:
\[ n = \frac{(1 \, \text{atm}) \times (492 \, \text{L})}{(0.082 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (300 \, \text{K})} \]

Calculating:
\[ n = \frac{492}{24.6} \approx 20 \, \text{moles of O₂} \]

### Step 2: Relate the moles of O₂ to the moles of NaClO₃.

From the balanced chemical equation:
\[ \text{NaClO}_3 + \text{Fe} \rightarrow \text{O}_2 + \text{FeO} + \text{NaCl} \]

It can be seen that 1 mole of NaClO₃ produces 1 mole of O₂. Therefore, the moles of NaClO₃ required will also be 20 moles.

### Step 3: Calculate the mass of NaClO₃ required.

The formula for calculating mass from moles is:
\[ \text{mass} = \text{moles} \times \text{molar mass} \]

The molar mass of NaClO₃ (sodium chlorate) can be calculated as follows:
- Na: 23 g/mol
- Cl: 35.5 g/mol
- O₃: 16 g/mol × 3 = 48 g/mol

Total molar mass of NaClO₃:
\[ 23 + 35.5 + 48 = 106.5 \, \text{g/mol} \]

Now, substituting the values:
\[ \text{mass} = 20 \, \text{moles} \times 106.5 \, \text{g/g mol} = 2130 \, \text{g} \]

### Step 4: Convert grams to kilograms.

To convert grams to kilograms:
\[ \text{mass in kg} = \frac{2130 \, \text{g}}{1000} = 2.13 \, \text{kg} \]

### Final Answer:
The mass of NaClO₃ required is **2.13 kg**.

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