Aluminum and $F e_{2} O_{3}$ are used as a fuel to produce energy according to given reaction
$2 A l (s) + F e_{2} O_{3} (s) \to A l_{2} O_{3} (s) + 2 F e (s)$
Given: $Δ H_{f}^{0} [A l_{2} O_{3}] = - 100$ $k c a l / m o l e$
$Δ H_{f}^{0} [F e_{2} O_{3}] = - 50$ $k c a l / m o l e$
Density of $A l = 2.7$ gram/c.c
Density of $F e_{2} O_{3} = 3.2$ gram/c.c
Select the correct statements

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Enthalpy change for the given reaction is – 50 k cal/mole

1.07 kg of mixture can produce maximum 200 k cal heat

Maximum calorific value of the fuel is 0.714 k cal/c.c

Maximum calorific value of the fuel is 0.432 k cal/gram

answer is A, D.

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Detailed Solution

A)    $Δ_{f}^{} H = Δ_{f}^{} H A l_{2} O_{3} - Δ_{f}^{} H F e_{2} O_{3}$
$\begin{array}{l} = - 100 - (- 50) \\ = - 50 \end{array}$
B) 1070 g mixture
wt. of AC in the mixture $\frac{54}{214} \times 1070 = 270$
moles of AC in the mixture $\frac{270}{54} = 5$
wt. of Fe2O3 in the mixture $\frac{160}{214} \times 1070 = 800$
moles of Fe2O3 in the mixture $\frac{800}{160} = 5$
$∴$ 1 mole $\to$ 50Kcal
5 mole $\to 50 \times 5 = 250$ Kcal
C) Calorific value Kcal/g
From balanced eqn a = no of moles
   $2 a \times 27 + a \times 160 = 1 \Rightarrow a = \frac{1}{214}$
Calorific value $\frac{1}{214} \times 50 = 0.233$ Kcal/g
D) Calorific value $\to$ Kcal/C.C
From balanced eqn a = no of moles
   $\begin{array}{l} 2 a \times \frac{27}{2.7} + a \times \frac{160}{3.2} = 1 \\ a = \frac{1}{70} \end{array}$
Calorific value $\frac{1}{70} \times 50 = 0.714$ Kcal/c.c