# How would you balance: copper sulfide + nitric acid --> Copper nitrate + sulfur + water + nitrogen monoxide?

Jan 21, 2016

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

#### Explanation:

Let $a , b , c , d , e , f$ be the number of molecules of each reactant. Th echecmical eqaution becomes

$a C u S + b H N {O}_{3} \to c C u {\left(N {O}_{3}\right)}_{2} + \mathrm{dS} + e {H}_{2} O + f N O$
Obsrvation reveals that nmber of molecules of $H N {O}_{3}$ must be an even number to accommodate ${H}_{2}$ on the right of the equation.

Let's start balancing for the radical $N {O}_{3}$.
On the right hand side of the equation the radical is either part of
atom $C u {\left(N {O}_{3}\right)}_{2}$ or can be obtained from combination of ${H}_{2} O$ and $N O$. We may choose lowest integer 1 for either and double the number of atoms of the other. However, choosing lowest integer 1 for $N O$ will rquire chosing 2 atoms of ${H}_{2} O$. This will increase the number of atoms of $H$ required for balancing. Therefore, choosing 2 atoms of $N O$ or $f = 2$, the equation becomes

$a C u S + b H N {O}_{3} \to c C u {\left(N {O}_{3}\right)}_{2} + \mathrm{dS} + e {H}_{2} O + 2 N O$

With this we need 3 more atoms of $O$ for the second atom of $N$. Implies $e = 4$. The equation becomes

$a C u S + b H N {O}_{3} \to c C u {\left(N {O}_{3}\right)}_{2} + \mathrm{dS} + 4 {H}_{2} O + 2 N O$
Now after having fixed number of atoms of $H$ on the right side, it follows that $b$ must$= 8$. To give us

$a C u S + 8 H N {O}_{3} \to c C u {\left(N {O}_{3}\right)}_{2} + \mathrm{dS} + 4 {H}_{2} O + 2 N O$
Now balacaing 24 number of $O$ atoms of left side with the right side atoms we obtain $c = 3$. This fixes $a = 3 = d$. We obtain the balanced equation as

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