# Question #13fbd

Nov 13, 2016

$\stackrel{0}{C} u + H \stackrel{+ 5}{N} {O}_{3} \to \stackrel{+ 2}{C} u {\left(N {O}_{3}\right)}_{2} + \stackrel{+ 2}{N} O + {H}_{2} O$
Change in Oxidation Number

$C u \left(0\right) \to C u \left(+ 2\right) \implies \text{2 unit oxidation}$

$N \left(+ 5\right) \to N \left(+ 2\right) \implies \text{3 unit reduction}$

$\text{Reacting Ratio of } C u \left(+ 2\right) : N \left(+ 5\right) = 3 : 2$

3mol $C u {\left(N {O}_{3}\right)}_{2}$ formed from 3mol Cu requires extra 6 mol $H N {O}_{3}$ as salt former which is added in LHS making total 8 mol $H N {O}_{3}$ in reactant side.The RHS is then adjusted as required.

$3 \stackrel{0}{C} u + 8 H \stackrel{+ 5}{N} {O}_{3} \to 3 \stackrel{+ 2}{C} u {\left(N {O}_{3}\right)}_{2} + 2 \stackrel{+ 2}{N} O + 4 {H}_{2} O$

Another Way

(1) First write the balanced equation for decomposition reaction of
$H N {O}_{3}$ forming the oxide of Nitrogen produced, water and oxygen as follows

$2 H N {O}_{3} \to {H}_{2} O + 2 N O + 3 O$

(2) Adjust the O-atom produced in step (1) combining with required proportion of Cu atoms as follows

$3 C u + 3 O \to 3 C u O$

(3) Write balanced equation of reaction of 3CuO with $H N {O}_{3}$ forming salt and water

$3 C u O + 6 H N {O}_{3} \to 3 C u {\left(N {O}_{3}\right)}_{2} + 3 {H}_{2} O$

(4) Finally add those equations to have the required balance equation.

$2 H N {O}_{3} \to {H}_{2} O + 2 N O + \cancel{3 O}$

$3 C u + \cancel{3 O} \to \cancel{3 C u O}$

$\cancel{3 C u O} + 6 H N {O}_{3} \to 3 C u {\left(N {O}_{3}\right)}_{2} + 3 {H}_{2} O$

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

Shortcut tips for balancing the equations of reactions between metal and nitric acids in various cases where oxides of nitrogen are produced.

(1) Let moles $C u$ be $x$ and moles of $H N {O}_{3}$ be $y$ in the balanced equation of a reaction.

Then $x = 5 - \text{ON of N in its oxide formed}$

As for above reaction $x = 5 - 2 = 3$

(2) Then $y = x \times v + v$,

$\text{where" v = "valency of metal or OS of metal in its nitrate salt }$

So for above reaction $v = 2$

and $y = 3 \times 2 + 2 = 8$

(3) Knowing the value of x and y we can easily balance the equation.