What mass of `MnO_(2)` is reduced by 35 mL of 0.16 N oxalic acid `(H_(2)C_(2)O_(4))` in acidic solution ? The skeleton equation is :-
`MnO_(2)+H^(+)+H_(2)C_(2)O_(4)toCO_(2)+H_(2)O+Mn^(2+)`

Options

A. 8.7 g
B. 0.24 g
C. 0.84 g
D. 43.5 g

Answer

To solve the problem of determining the mass of \( \text{MnO}_2 \) reduced by 35 mL of 0.16 N oxalic acid in acidic solution, we can follow these steps:

### Step 1: Write the Skeleton Equation
The skeleton equation is given as:
\[ \text{MnO}_2 + \text{H}^+ + \text{H}_2\text{C}_2\text{O}_4 \rightarrow \text{CO}_2 + \text{H}_2\text{O} + \text{Mn}^{2+} \]

### Step 2: Balance the Equation
To balance the equation, we first balance the carbon atoms by placing a coefficient of 2 in front of \( \text{H}_2\text{C}_2\text{O}_4 \):
\[ \text{MnO}_2 + \text{H}^+ + 2\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O} + \text{Mn}^{2+} \]
Next, we balance the hydrogen atoms. The total number of hydrogen atoms on the left side is 4 (from 2 \( \text{H}_2\text{C}_2\text{O}_4 \)) plus the protons \( \text{H}^+ \). We can balance it by adjusting the coefficient of \( \text{H}^+ \) accordingly.

After balancing, we find:
\[ \text{MnO}_2 + 4\text{H}^+ + 2\text{H}_2\text{C}_2\text{O}_4 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O} + \text{Mn}^{2+} \]

### Step 3: Calculate the n-factor of Oxalic Acid
The n-factor of oxalic acid \( \text{H}_2\text{C}_2\text{O}_4 \) is 2 because it can donate 2 protons (or electrons) in this reaction.

### Step 4: Calculate Molarity of Oxalic Acid
Given the normality (N) of oxalic acid is 0.16 N, we can calculate molarity (M) using the formula:
\[ \text{Molarity} = \frac{\text{Normality}}{\text{n-factor}} = \frac{0.16}{2} = 0.08 \, \text{mol/L} \]

### Step 5: Calculate Moles of Oxalic Acid
To find the moles of oxalic acid, we use the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (in L)} \]
Converting 35 mL to liters:
\[ \text{Volume} = 35 \, \text{mL} = 35 \times 10^{-3} \, \text{L} \]
Now, calculate moles:
\[ \text{Moles of } \text{H}_2\text{C}_2\text{O}_4 = 0.08 \, \text{mol/L} \times 35 \times 10^{-3} \, \text{L} = 2.8 \times 10^{-3} \, \text{mol} \]

### Step 6: Determine Moles of \( \text{MnO}_2 \)
From the balanced equation, we see that 1 mole of \( \text{MnO}_2 \) reacts with 2 moles of oxalic acid. Therefore, the moles of \( \text{MnO}_2 \) will also be:
\[ \text{Moles of } \text{MnO}_2 = 2.8 \times 10^{-3} \, \text{mol} \]

### Step 7: Calculate Mass of \( \text{MnO}_2 \)
The molar mass of \( \text{MnO}_2 \) is approximately 87 g/mol. We can now calculate the mass:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} = 2.8 \times 10^{-3} \, \text{mol} \times 87 \, \text{g/mol} = 0.2436 \, \text{g} \]
Rounding this to two decimal places gives us:
\[ \text{Mass} \approx 0.24 \, \text{g} \]

### Final Answer
The mass of \( \text{MnO}_2 \) reduced by 35 mL of 0.16 N oxalic acid is approximately **0.24 g**.

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