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In the thermite reaction used for obtaining energy aluminium and ferric oxide react to form iron and aluminium oxide. From the given data identify the option(s) which is(are) correct :
[Given ${ΔH}_{f}^{\circ} {Al}_{2} O_{3} = - 390 kJ / mol$ , ${ΔH}_{f}^{\circ} {Fe}_{2} O_{3} = - 176 kJ / mol,$
density of alumintun1 $= 2.7 gm / ml$ , density of ${Fe}_{2} O_{3} = 3.2 gm / ml$ ]

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$140 ml$ of the mixture can produce $280$ $KJ$ of heat.

Maximum calorific value of the fuel can be $1000 J / gm$

Maximum calorific value of the fuel can be $\frac{21.4}{7}$ $kJ / ml$

$2.14 Kg$ of the mixture can produce $2500 KJ$ of heat.

answer is A, B, D.

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Detailed Solution

$\begin{array}{l} 2 Al + {Fe}_{2} O_{3} ⟶ {Al}_{2} O_{3} + 2 Fe \\ {ΔH}_{Reaction}^{\circ} = - 390 + 176 = - 214 kJ / mol \end{array}$

(A) $2 a \times 27 + a \times 160 = 1$

$a = \frac{1}{214}$

Calorific value= $e = 214 \times \frac{1}{214} \times 1000 J / g$

(b) $\frac{2 b \times 27}{2.7} + \frac{b \times 160}{3.2} = 1 \Rightarrow b = \frac{1}{70}$

Calorific value $= 214 \times \frac{1}{70} kJ / mL$

(d) $\frac{140}{70} \times 214 kJ = 428 kJ$ (maximum)

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