The amount of Mg in gms. to be dissolved in dilute H₂SO₄ to liberate H₂ which is just sufficient to reduce 160g of ferric oxide is

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answer is C.

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Detailed Solution

To solve this, we follow these steps:

Identify the reaction:
Magnesium reacts with sulfuric acid:
Mg + H₂SO₄ → MgSO₄ + H₂
The liberated hydrogen then reduces ferric oxide:
Fe₂O₃ + 3H₂ → 2Fe + 3H₂O
Use stoichiometry:
The molar mass of Fe₂O₃ = 2 × 55.8 + 3 × 16 = 159.6 g/mol.
One mole of Fe₂O₃ requires three moles of H₂.
Given: 160 g of Fe₂O₃ is approximately 1 mole (160 g ≈ 159.6 g).
Thus, 1 mole of Fe₂O₃ requires 3 moles of H₂.
Calculate the amount of Mg:
From the first reaction, 1 mole of Mg liberates 1 mole of H₂.
Therefore, 3 moles of H₂ require 3 moles of Mg.
The molar mass of Mg is 24 g/mol.
Hence, 3 moles of Mg weigh 3 × 24 = 72 g.

Answer: (c) 72

$\large \mathop {Mg}\limits^{1\;mole} \; + \;{H_2}S{O_4}\; \to \;MgS{O_4}\; + \;\mathop {{H_2}}\limits^{1\;mole}$

$\large \mathop {F{e_2}{O_3}}\limits_{160g}^{1\;mole} + \;\;\mathop {3{H_2}}\limits_{3\;mole}^{3\;mole} \; \to \;2Fe\; + \;3{H_2}O$

$\large \\1\;mole\;of\;Mg\; \equiv \;1\;mole\;of\;{H_2} \\'X'\;g\;of\;Mg\; \equiv \;3\;mole\;of{H_2}$

$\large X\; = \;\frac{{3 \times 24}}{1}\; = \;72\;g$

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