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 step by step, we will first calculate the mass of \( K_2Cr_2O_7 \) required to produce 5.0 L of \( CO_2 \) at 77°C and 0.82 atm pressure, and then we will determine the volume of 0.1 N NaOH required to neutralize the \( CO_2 \) evolved.

### Step 1: Calculate the number of moles of \( CO_2 \)

We can use the Ideal Gas Law equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure in atm (0.82 atm)
- \( V \) = volume in liters (5.0 L)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin (77°C + 273 = 350 K)

Rearranging the equation to find \( n \):

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

Substituting the values:

\[ n = \frac{(0.82 \, \text{atm}) \times (5.0 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (350 \, \text{K})} \]

Calculating \( n \):

\[ n = \frac{4.1}{28.735} \approx 0.142 \, \text{moles of } CO_2 \]

### Step 2: Determine the stoichiometry of the reaction

The balanced reaction between \( K_2Cr_2O_7 \) and oxalic acid \( (H_2C_2O_4) \) is:

\[ K_2Cr_2O_7 + 7H_2C_2O_4 \rightarrow 2K_2C_2O_4 + 6CO_2 + 7H_2O \]

From the balanced equation, we see that 1 mole of \( K_2Cr_2O_7 \) produces 6 moles of \( CO_2 \).

### Step 3: Calculate the moles of \( K_2Cr_2O_7 \) required

Using the stoichiometry:

\[ \text{For } 6 \text{ moles of } CO_2, \text{ 1 mole of } K_2Cr_2O_7 \text{ is required.} \]

Thus, for 0.142 moles of \( CO_2 \):

\[ \text{Moles of } K_2Cr_2O_7 = \frac{0.142}{6} \approx 0.02367 \, \text{moles} \]

### Step 4: Calculate the mass of \( K_2Cr_2O_7 \)

The molar mass of \( K_2Cr_2O_7 \) is approximately 294 g/mol. Therefore, the mass required is:

\[ \text{Mass} = \text{moles} \times \text{molar mass} = 0.02367 \, \text{moles} \times 294 \, \text{g/mol} \approx 6.95 \, \text{grams} \]

### Step 5: Calculate the volume of 0.1 N NaOH required to neutralize \( CO_2 \)

The \( CO_2 \) produced will react with NaOH. The reaction is:

\[ CO_2 + 2NaOH \rightarrow Na_2CO_3 + H_2O \]

From the reaction, 1 mole of \( CO_2 \) reacts with 2 equivalents of NaOH. Therefore, the number of equivalents of \( CO_2 \) produced is:

\[ \text{Equivalents of } CO_2 = 0.142 \, \text{moles} \times 2 = 0.284 \, \text{equivalents} \]

Now, using the normality of NaOH (0.1 N):

\[ \text{Volume of NaOH} = \frac{\text{equivalents}}{\text{normality}} = \frac{0.284 \, \text{equivalents}}{0.1 \, \text{N}} = 2.84 \, \text{L} \]

### Final Answers

1. Mass of \( K_2Cr_2O_7 \) required: **7 grams**
2. Volume of 0.1 N NaOH required: **2.84 liters**