Mg s - f -2 O 2 g - MgO s H -140-2 kJ - -2What is H of the reaction- 3 Mg - F e 2 O 3- 3 MgO -2 FeA- -272-3 kJB- 272-3 kJC- 272-2 kJD- -227-2 kJ

The correct option is D 227.2 kJ Equation (2) is multiplied by 3 2Fe(s) + 3/2 O_2(g) → Fe_2O_3(s) H = 193.4 kJ. . . . . . (1) 3Mg(s) + 3/2 O_2(g) ...

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Selecting Consecutive Terms in GP

Mg s + f /2 O...

Question

$2 F e (s) + \frac{3}{2} O_{2} (g) \to F e_{2} O_{3} (s) (△ H = - 193.4 k J) . . . . . . (1)$
$M g (s) + \frac{1}{2} O_{2} (g) \to M g O (s) (△ H = - 140.2 k J) . . . . . . (2)$

What is $△$ H of the reaction? $3 M g + F e_{2} O_{3} \to 3 M g O + 2 F e$

-272.3 kJ

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272.3 kJ

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272.2 kJ

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-227.2 kJ

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Similar questions

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

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)

Δ H

3 M g + {F e}_{2} O_{3} ⟶ 3 M g O + 2 F e

m

F e_{2} O_{3} (s) \to F e (s) + 3 / 2 O_{2} (g)

3 M g + N_{2} ⟶ M g_{3} N_{2}

M g

10

M g_{3} N_{2}

100 %

x

x \times 100

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