Question

How would you balance:
$copper{\text{ }}sulfide{\text{ }} + {\text{ }}nitric{\text{ }}acid\; \to \;\;copper{\text{ }}nitrate{\text{ }} + {\text{ }}sulfur{\text{ }} + {\text{ }}water{\text{ }} + {\text{ }}nitrogen{\text{ }}monoxide$ ?

Answer 1

In a balancing of a chemical equation, the stoichiometric coefficients are added to the reactants and products. The chemical equation must follow the law of conservation of mass and the law of constant proportions. i.e., the same number of atoms of each element should be present on the reactant side and the product side of the equation.

Complete answer:
Let’s consider a, b, c, d, e and f be the number of molecules of each reactant. The chemical equation becomes
$aCuS + bHN{O_3} \to cCu{\left( {N{O_3}} \right)_2} + dS + e{H_2}O + fNO$
Observation reveals that the number of molecules of $HN{O_3}$ 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 $Cu{\left( {N{O_3}} \right)_{2\;}}$ or can be obtained from combination of ${H_{2\;}}O$ and $NO$ . We may choose the lowest integer $1$ for either and double the number of atoms of the other. However, choosing the lowest integer $1$ for $NO$ will require choosing $2$ atoms of ${H_{2\;}}O$ . This will increase the number of atoms of $H$ required for balancing. Therefore, choosing $2$ atoms of $NO$ or $f = 2$ , the equation becomes
$aCuS + bHN{O_3} \to cCu{\left( {N{O_3}} \right)_2} + dS + e{H_2}O + 2NO$
With this we need $3$ more atoms of $O$ for the second atom of $N$ . Implies $e = 4$ . The equation becomes
$aCuS + bHN{O_3} \to cCu{\left( {N{O_3}} \right)_2} + dS + 4{H_2}O + 2NO$
Now after having fixed the number of atoms of $H$ on the right side, it follows that $b$ must be $8$ . To give us
$aCuS + 8HN{O_3} \to cCu{\left( {N{O_3}} \right)_2} + dS + 4{H_2}O + 2NO$
Now balancing the$24$ number of $O$ atoms of the left side with the right-side atoms we obtain $c = 3$. This fix $a = 3 = d$ .
We obtain the balanced equation as;
$\;3CuS + 8HN{O_3} \to 3Cu{\left( {N{O_3}} \right)_2} + 3S + 4{H_2}O + 2NO$

Note: There are two types of balancing the chemical equation. The first is a traditional method and the second is an algebraic method. The algebraic method is easy and convenient to use. In an algebraic method, the balancing of chemical equations involves assigning algebraic variables as stoichiometric coefficients to reactants and products in the unbalanced chemical equation. These variables are used in mathematical equations.
$aCuS + bHN{O_3} \to cCu{\left( {N{O_3}} \right)_2} + dS + e{H_2}O + fNO$
After solving we obtain the equation
$\;3CuS + 8HN{O_3} \to 3Cu{\left( {N{O_3}} \right)_2} + 3S + 4{H_2}O + 2NO$