The mass of `K_(2)Cr_(2)O_(7)` required to produce 5.0 L `CO_(2)` at `77^(@)C` and 0.82 atm pressure from excess of oxalic acid and volume of 0.1 N `NaOH` required to neutralise the `CO_(2)` evolved respectively are

Options

A. `7 g, 2.86 L`
B. `5 g, 1.86L`
C. `4 g, 0.86L`
D. none

Explanation

To solve the problem, we need to determine the mass of potassium dichromate (K₂Cr₂O₇) required to produce 5.0 L of CO₂ at 77°C and 0.82 atm pressure, and the volume of 0.1 N NaOH required to neutralize the CO₂ evolved.

### Step 1: Calculate the number of moles of CO₂ produced
We can use the ideal gas law to find the number of moles of CO₂ produced. The ideal gas law is given by:

\[ PV = nRT \]

Where:
- \( P \) = pressure in atm (0.82 atm)
- \( V \) = volume in liters (5.0 L)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin (77°C = 77 + 273.15 = 350.15 K)

Rearranging the ideal gas law to solve for \( n \):

\[ n = \frac{PV}{RT} \]

Substituting the values:

\[ n = \frac{(0.82 \, \text{atm}) \times (5.0 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (350.15 \, \text{K})} \]

Calculating \( n \):

\[ n = \frac{4.1}{28.703} \approx 0.143 \, \text{moles of CO₂} \]

### Step 2: Determine the moles of K₂Cr₂O₇ required
The oxidation of oxalic acid (H₂C₂O₄) to CO₂ requires K₂Cr₂O₇ as an oxidizing agent. The balanced reaction shows that 1 mole of K₂Cr₂O₇ produces 3 moles of CO₂.

Using the stoichiometry of the reaction:

\[ \text{Moles of K₂Cr₂O₇} = \frac{1}{3} \times \text{Moles of CO₂} \]

Substituting the moles of CO₂:

\[ \text{Moles of K₂Cr₂O₇} = \frac{1}{3} \times 0.143 \approx 0.0477 \, \text{moles of K₂Cr₂O₇} \]

### Step 3: Calculate the mass of K₂Cr₂O₇ required
To find the mass, we use the formula:

\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]

The molar mass of K₂Cr₂O₇ is approximately 294 g/mol.

Calculating the mass:

\[ \text{Mass of K₂Cr₂O₇} = 0.0477 \, \text{moles} \times 294 \, \text{g/mol} \approx 14.05 \, \text{g} \]

### Step 4: Calculate the volume of 0.1 N NaOH required
The CO₂ produced will react with NaOH in a 1:1 ratio. Therefore, the moles of NaOH required will be equal to the moles of CO₂ produced.

Since we have 0.143 moles of CO₂, we need 0.143 moles of NaOH.

Using the normality formula:

\[ \text{Normality (N)} = \frac{\text{equivalents}}{\text{volume in L}} \]

For NaOH, which has one hydroxide ion per molecule, the equivalents are equal to the moles.

Thus, we can rearrange to find the volume:

\[ V = \frac{\text{moles}}{\text{Normality}} \]

Substituting the values:

\[ V = \frac{0.143 \, \text{moles}}{0.1 \, \text{N}} = 1.43 \, \text{L} \]

### Final Results
- Mass of K₂Cr₂O₇ required: **14.05 g**
- Volume of 0.1 N NaOH required: **1.43 L**