The equivalent weight of $H N O_{3}$ (molecular weight = $63$ ) in the following reaction is:

$3 C u + 8 H N O_{3} \to 3 C u (N O_{3})_{2} + 2 N O + 4 H_{2} O$

\frac{4 \times 63}{3}

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\frac{63}{5}

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\frac{63}{3}

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\frac{63}{8}

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Solution

The correct option is B $\frac{4 \times 63}{3}$
The reaction is
$3 C u + 8 {H N O}_{3} + 5 ⟶ 3 C u {({N O}_{3})}_{2} + 2 N O + 2 + 4 H_{2} O$
change in electrons $= 3$
molecular weight $= 63$
Equivalent weight $= \frac{4 \times 63}{3}$
Ans :- Option A.

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Similar questions

{H N O}_{3}

3 C u + 8 H N O_{3} ⟶ 3 C u {({N O}_{3})}_{2} + 2 N O + 4 H_{2} O

3 C u + 8 H N O_{3} \to 3 C u (N O_{3})_{2} + 4 H_{2} O + 2 N O

H N O_{3}

H N O_{3} = 63) .

3 C u + 8 H N O_{3} \to 3 C u (N O_{3})_{2} + 2 N O + 4 H_{2} O

H_{2} O

10

H N O_{3}

10

C u

3 C u + 8 H N O_{3} ⟶ 3 C u (N O_{3})_{2} + 2 N O + 4 H_{2} O

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