How many grams of `H_(2)O_(2)` are required to convert 0.1 moles PbS to 0.1 moles `PbSO_(4)`-

Options

A. 12.8g
B. 13.6g
C. 16g
D. 3.4g

Answer

To solve the problem of how many grams of \( H_2O_2 \) are required to convert 0.1 moles of \( PbS \) to 0.1 moles of \( PbSO_4 \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between lead sulfide (\( PbS \)) and hydrogen peroxide (\( H_2O_2 \)) can be represented as follows:

\[ PbS + H_2O_2 \rightarrow PbSO_4 + H_2O \]

### Step 2: Balance the chemical equation
To balance the equation, we need to ensure that the number of atoms for each element is the same on both sides of the equation. The balanced equation is:

\[ PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O \]

This indicates that 1 mole of \( PbS \) reacts with 4 moles of \( H_2O_2 \) to produce 1 mole of \( PbSO_4 \) and 4 moles of water.

### Step 3: Determine the moles of \( H_2O_2 \) required
From the balanced equation, we see that to convert 1 mole of \( PbS \) to 1 mole of \( PbSO_4 \), we need 4 moles of \( H_2O_2 \). Therefore, for 0.1 moles of \( PbS \):

\[ \text{Moles of } H_2O_2 = 0.1 \text{ moles of } PbS \times 4 = 0.4 \text{ moles of } H_2O_2 \]

### Step 4: Calculate the mass of \( H_2O_2 \)
To find the mass of \( H_2O_2 \), we use the formula:

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

The molar mass of \( H_2O_2 \) is calculated as follows:

\[ \text{Molar Mass of } H_2O_2 = 2 \times 1 + 32 = 34 \text{ g/mol} \]

Now, substituting the values:

\[ \text{Mass of } H_2O_2 = 0.4 \text{ moles} \times 34 \text{ g/mol} = 13.6 \text{ grams} \]

### Final Answer
Thus, the mass of \( H_2O_2 \) required to convert 0.1 moles of \( PbS \) to 0.1 moles of \( PbSO_4 \) is **13.6 grams**.

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