In the reaction
`aFe(s) + bH_(2)O(g) to mFe_(3)O_(4)(s) + nH_(2)(g)`
a, b, m and n respectively are

Options

A. 6, 8, 2, 8,
B. 3, 4, 1, 4
C. 3, 4, 4, 1
D. 2, 3, 4, 4

Answer

To balance the chemical reaction \( aFe(s) + bH_2O(g) \rightarrow mFe_3O_4(s) + nH_2(g) \), we will follow these steps:

### Step 1: Write down the unbalanced equation
The unbalanced equation is:
\[ Fe(s) + H_2O(g) \rightarrow Fe_3O_4(s) + H_2(g) \]

### Step 2: Count the number of atoms of each element on both sides
- **Reactants:**
- Iron (Fe): 1
- Oxygen (O): 1
- Hydrogen (H): 2

- **Products:**
- Iron (Fe): 3 (from \( Fe_3O_4 \))
- Oxygen (O): 4 (from \( Fe_3O_4 \))
- Hydrogen (H): 2 (from \( H_2 \))

### Step 3: Balance the iron (Fe) atoms
Since there are 3 iron atoms in \( Fe_3O_4 \) and only 1 in the reactants, we need to multiply the iron in the reactants by 3:
\[ 3Fe(s) + H_2O(g) \rightarrow Fe_3O_4(s) + H_2(g) \]

### Step 4: Balance the oxygen (O) atoms
Now, we have 4 oxygen atoms in \( Fe_3O_4 \) on the product side. Since water (\( H_2O \)) contains 1 oxygen atom, we need 4 water molecules:
\[ 3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + H_2(g) \]

### Step 5: Balance the hydrogen (H) atoms
Now, we have 8 hydrogen atoms from 4 water molecules on the reactant side. The product side has 2 hydrogen atoms from \( H_2 \). To balance the hydrogen, we need 4 \( H_2 \) molecules:
\[ 3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + 4H_2(g) \]

### Step 6: Final balanced equation
The final balanced equation is:
\[ 3Fe(s) + 4H_2O(g) \rightarrow 1Fe_3O_4(s) + 4H_2(g) \]

### Step 7: Identify the coefficients
From the balanced equation, we can identify:
- \( a = 3 \)
- \( b = 4 \)
- \( m = 1 \)
- \( n = 4 \)

### Summary of values:
- \( a = 3 \)
- \( b = 4 \)
- \( m = 1 \)
- \( n = 4 \)

---