The balanced equation for the reaction of dilute nitric acid with copper is :

3 C u + 8 {H N O}_{3} (d i l .) \to 3 C u {({N O}_{3})}_{2} + 4 H_{2} O + 2 N O

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C u + 8 {H N O}_{3} (d i l .) \to C u {({N O}_{3})}_{2} + 4 H_{2} O + 2 N O

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3 C u + 4 {H N O}_{3} (d i l .) \to 3 C u {({N O}_{3})}_{2} + 2 H_{2} O + 2 N O

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C u + 8 {H N O}_{3} (d i l .) \to C u {({N O}_{3})}_{2} + 2 H_{2} O + N O

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Solution

The correct option is A $3 C u + 8 {H N O}_{3} (d i l .) \to 3 C u {({N O}_{3})}_{2} + 4 H_{2} O + 2 N O$
$3 C u + 8 {H N O}_{3} (d i l .) \to 3 C u {({N O}_{3})}_{2} + 4 H_{2} O + 2 N O$

This is the correct chemical equation as it follows the law of conservation of mass.

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Similar questions

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

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

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

{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

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