$4 M g + 10 {H N O}_{3} ⟶ 4 M g {({N O}_{3})}_{2} + {N H}_{4} {N O}_{3} + 3 H_{2} O$
The amount of magnesium required to reduce 1 mole of ${H N O}_{3}$ in the reaction is

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$96 g$

$48 g$

$9.6 g$

$4.8 g$

answer is A.

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Detailed Solution

The oxidation state of nitrogen in ${H N O}_{3}$ is $+ 5$

The oxidation state of nitrogen in $M g {({N O}_{3})}_{2}$ is $+ 5$

The oxidation state of nitrogen in ${N H}_{4} {N O}_{3}$ is $- 3$ and $+ 5$

Only one nitrogen out of ten gets reduced

Four moles of magnesium reduces one mole of ${H N O}_{3}$

Atomic weight of magnesium is 24

Therefore, $4 \times 24 = 1$ mole nitric acid

$96 = 1$ mole nitric acid

Hence the correct option is A.

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