The volume of `H_(2)` STP required to completely reduce 160 gms of`Fe_(2)O_(3)` is

Options

A. `3xx22.4L`
B. `2xx22.4L`
C. `22.4L`
D. `11.2 L`

Explanation

To solve the problem of determining the volume of \( H_2 \) at STP required to completely reduce 160 g of \( Fe_2O_3 \), we will follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The reduction of \( Fe_2O_3 \) by \( H_2 \) can be represented by the following balanced equation:
\[ Fe_2O_3 + 3H_2 \rightarrow 2Fe + 3H_2O \]

2. **Calculate the Molar Mass of \( Fe_2O_3 \)**:
The molar mass of \( Fe_2O_3 \) can be calculated as follows:
- Iron (Fe) has a molar mass of approximately 56 g/mol.
- Oxygen (O) has a molar mass of approximately 16 g/mol.
\[ \text{Molar mass of } Fe_2O_3 = (2 \times 56) + (3 \times 16) = 112 + 48 = 160 \text{ g/mol} \]

3. **Calculate the Number of Moles of \( Fe_2O_3 \)**:
Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{160 \text{ g}}{160 \text{ g/mol}} = 1 \text{ mole} \]

4. **Determine the Moles of \( H_2 \) Required**:
From the balanced equation, we see that 1 mole of \( Fe_2O_3 \) reacts with 3 moles of \( H_2 \):
\[ \text{Moles of } H_2 = 3 \times \text{Moles of } Fe_2O_3 = 3 \times 1 = 3 \text{ moles} \]

5. **Calculate the Volume of \( H_2 \) at STP**:
At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters. Therefore, the volume of \( H_2 \) required is:
\[ \text{Volume of } H_2 = \text{Moles of } H_2 \times 22.4 \text{ L/mol} = 3 \times 22.4 \text{ L} = 67.2 \text{ L} \]

### Final Answer:
The volume of \( H_2 \) required to completely reduce 160 g of \( Fe_2O_3 \) at STP is **67.2 liters**.

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