Equal masses of iron and magnesium, in separate reactions with dilute $H_{2} S O_{4}$ , gave 'x' g of hydrogen and 'y' g of hydrogen respectively. 'y' g of hydrogen is just sufficient to reduce $0.75 m o l e s o f F e_{3} O_{4}$ . Value of 'x' and 'y' is:

x - 1.287 gm , y - 6 gm

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x - 2.57 gm , y - 3gm

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x - 1.287 gm , y - 3 gm

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x - 2.57 gm , y - 6 gm

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Solution

The correct option is D x - 2.57 gm , y - 6 gm
$F e_{3} O_{4} + 4 H_{2} ⟶ 3 F e + 4 H_{2} O$
$1 m o l e F e_{3} O_{4} ⟶ 4 m o l e s H_{2}$
$0.75 m o l e F e_{3} O_{4} ⟶ 4 \times 0.75 = 3 m o l e s$
$F e + H_{2} S O_{4} ⟶ F e S O_{4} + H_{2}$
$M g + H_{2} S O_{4} ⟶ M g S O_{4} + H_{2}$
Hence, 3 moles of $H_{2}$ required for reaction with Mg.
$3 m o l e s o f M g ⟶ 72 g$
$56 g F e ⟶ 1 m o l e H_{2}$
$72 g F e ⟶ \frac{72}{56} = 1.28 m o l e s$
Hence, 1.28 moles of $H_{2}$ required for Fe.
$x = 1.28 \times 2 = 2.57 g$
$y = 3 \times 2 = 6 g$

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