An intimate mixture of `Fe_(2)O_(3)` and Ai is used in solid rocket fuel. Calculate the fuel value per gram of mixture. `DeltaH_(Al_(2)O_(3)) = 399.0 kcal, DeltaH_(Fe_(2)O_(3))=199.0` kcal, density of `Fe_(2)O_(3)` and Al are 5.2 g/ml and 2.7 g/ml respectively.

Options

A. 39.2Kcal
B. 0.392Kcal
C. 3.92Kcal
D. 392Kcal

Answer

To solve the problem of calculating the fuel value per gram of the mixture of `Fe2O3` and `Al`, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between aluminum and iron(III) oxide can be written as:
\[ 4Al + 3Fe_2O_3 \rightarrow 2Al_2O_3 + 6Fe \]

### Step 2: Determine the change in enthalpy (ΔH) for the reaction
Using the given data:
- \( \Delta H_{Al_2O_3} = -399.0 \, \text{kcal/mol} \) (since it is the formation of aluminum oxide)
- \( \Delta H_{Fe_2O_3} = -199.0 \, \text{kcal/mol} \)

The change in enthalpy for the reaction can be calculated as:
\[ \Delta H_{reaction} = \text{(enthalpy of products)} - \text{(enthalpy of reactants)} \]
\[ \Delta H_{reaction} = [2 \times \Delta H_{Al_2O_3}] - [4 \times \Delta H_{Al} + 3 \times \Delta H_{Fe_2O_3}] \]
Since \( \Delta H_{Al} = 0 \) (as it is in its elemental state), we have:
\[ \Delta H_{reaction} = [2 \times (-399.0)] - [3 \times (-199.0)] \]
Calculating this gives:
\[ \Delta H_{reaction} = -798 + 597 = -201 \, \text{kcal} \]

### Step 3: Calculate the molar masses
- Molar mass of `Al` = 27 g/mol
- Molar mass of `Fe2O3` = 160 g/mol

### Step 4: Calculate the density and volume of the mixture
Using the densities provided:
- Density of `Fe2O3` = 5.2 g/ml
- Density of `Al` = 2.7 g/ml

Let’s assume we have 1 mole of `Fe2O3` and 4/3 moles of `Al` (from the balanced equation):
- Mass of `Fe2O3` = 160 g
- Mass of `Al` = \( 4 \times 27 = 108 \, g \)

Total mass of the mixture = \( 160 + 108 = 268 \, g \)

Now, calculate the volume:
- Volume of `Fe2O3` = \( \frac{160 \, g}{5.2 \, g/ml} \approx 30.77 \, ml \)
- Volume of `Al` = \( \frac{108 \, g}{2.7 \, g/ml} \approx 40.0 \, ml \)

Total volume of the mixture = \( 30.77 + 40.0 \approx 70.77 \, ml \)

### Step 5: Calculate the fuel value per gram of the mixture
The fuel value is given as:
\[ \text{Fuel value} = \frac{\Delta H_{reaction}}{\text{Total mass of the mixture}} = \frac{-201 \, \text{kcal}}{268 \, g} \]
Calculating this gives:
\[ \text{Fuel value} \approx -0.749 \, \text{kcal/g} \]

### Final Answer
The fuel value per gram of the mixture is approximately \( 0.749 \, \text{kcal/g} \).

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