Given that- 2Fes- 3 2 O 2 g- Fe 2 O 3 s- H-193-4kJ-i Mgs- 1 2 O 2 g- MgOs- H-140-2kJ-iiWhat is - H of the reaction- 3Mg- Fe 2 O 3 - 3MgO-2Fe

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Hess' Law

Given that: ...

Question

Given that:
$2 F e (s) + \frac{3}{2} O_{2} (g) ⟶ {F e}_{2} O_{3} (s); Δ H = - 193.4 k J . . . (i)$
$M g (s) + \frac{1}{2} O_{2} (g) ⟶ M g O (s); Δ H = - 140.2 k J . . . . (i i)$
What is $Δ H$ of the reaction?
$3 M g + {F e}_{2} O_{3} ⟶ 3 M g O + 2 F e$

- 227.2 k J

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- 237.3 k J

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2227.2 k J

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- 257.3 k J

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2 F e (s) + \frac{3}{2} O_{2} (g) \to F e_{2} O_{3} (s) (△ H = - 193.4 k J) . . . . . . (1)

M g (s) + \frac{1}{2} O_{2} (g) \to M g O (s) (△ H = - 140.2 k J) . . . . . . (2)

△

3 M g + F e_{2} O_{3} \to 3 M g O + 2 F e

Δ H

85^{o} C

{F e}_{2} O_{3} (s) + 3 H_{2} ⟶ 2 F e (s) + 3 H_{2} O (l)

Δ H_{298}^{o} = - 33.29 k J / m o l

Substance	${F e}_{2} O_{3} (s)$	$F e (s)$	$H_{2} O (l)$	$H_{2} (g)$
$C_{P}^{o} (J / K m o l)$	$103.8$	$25.1$	$75.3$	$28.8$

2 F e (s) + \frac{3}{2} O_{2} (g) \to F e_{2} O_{3} (s) (△ H = - 193.4 k J) . . . . . . (1)

M g (s) + \frac{1}{2} O_{2} (g) \to M g O (s) (△ H = - 140.2 k J) . . . . . . (2)

△

3 M g + F e_{2} O_{3} \to 3 M g O + 2 F e

{F e}_{2} O_{3} + 3 C O \to 2 F e + 3 {C O}_{2} F e O + C O \to F e + {C O}_{2} △ r_{H} = - 26 k J △ r_{H} = - 16 k J

△ H

{F e}_{2} O_{3} + C O \to 2 F e O + {C O}_{2} △ r_{H} = ?

{F e}_{2} O_{3} + 2 A l \to {A l}_{2} O_{3} + 2 F e

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