`Fe_(2)O_(3)(s)` may be converted to Fe by the reaction
`Fe_(2)O_(3)(s)+3H_(2)(g)hArr2Fe(s)+3H_(2)O(g)` for which `K_(c)=8 `at temp . `720^(@)c`.
What percentage of the `H_(2)` ramains unreacted after the reaction hascome to equilibrium ?

Options

A. `22%`
B. `34%`
C. `66%`
D. `78%`

Answer

To solve the problem, we need to determine the percentage of unreacted hydrogen after the reaction reaches equilibrium. The reaction is given as:

\[ \text{Fe}_2\text{O}_3(s) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{Fe}(s) + 3 \text{H}_2\text{O}(g) \]

with an equilibrium constant \( K_c = 8 \) at a temperature of \( 720^\circ C \).

### Step-by-Step Solution:

1. **Write the Expression for \( K_c \)**:
The equilibrium constant \( K_c \) for the reaction can be expressed as:
\[ K_c = \frac{[\text{H}_2\text{O}]^3}{[\text{H}_2]^3} \]
Here, we do not include solids in the expression.

2. **Set Up Initial Concentrations**:
Let's assume we start with 1 mole of \( \text{H}_2 \) and no products. At the beginning:
- \([\text{H}_2] = 1 \, \text{mol}\)
- \([\text{H}_2\text{O}] = 0 \, \text{mol}\)

3. **Define Change at Equilibrium**:
Let \( x \) be the amount of \( \text{H}_2 \) that reacts. The changes in concentrations will be:
- \([\text{H}_2] = 1 - 3x\)
- \([\text{H}_2\text{O}] = 3x\)

4. **Substitute into the \( K_c \) Expression**:
Substitute the equilibrium concentrations into the \( K_c \) expression:
\[ 8 = \frac{(3x)^3}{(1 - 3x)^3} \]
Simplifying gives:
\[ 8 = \frac{27x^3}{(1 - 3x)^3} \]

5. **Cross-Multiply and Simplify**:
Cross-multiplying leads to:
\[ 8(1 - 3x)^3 = 27x^3 \]
Expanding and simplifying this equation will help us find \( x \).

6. **Solve for \( x \)**:
To simplify the calculations, we can take the cube root of both sides:
\[ 2(1 - 3x) = 3x \]
Rearranging gives:
\[ 2 - 6x = 3x \implies 2 = 9x \implies x = \frac{2}{9} \]

7. **Calculate Unreacted Hydrogen**:
The amount of \( \text{H}_2 \) that remains unreacted is:
\[ [\text{H}_2]_{\text{unreacted}} = 1 - 3x = 1 - 3\left(\frac{2}{9}\right) = 1 - \frac{6}{9} = \frac{3}{9} = \frac{1}{3} \]

8. **Calculate the Percentage of Unreacted Hydrogen**:
The percentage of unreacted \( \text{H}_2 \) is:
\[ \text{Percentage} = \left(\frac{[\text{H}_2]_{\text{unreacted}}}{[\text{H}_2]_{\text{initial}}}\right) \times 100 = \left(\frac{\frac{1}{3}}{1}\right) \times 100 = 33.33\% \]

### Final Answer:
Approximately **34%** of the hydrogen remains unreacted after the reaction has come to equilibrium.