Which is the next logical step in balancing the given equation

CS2 (l) + Cl2 (g) = CCl4 (l) + S2Cl2 (l)

A. Place the coefficient 2 in front of sulfur dichloride (S2Cl2)

B. Place the coefficient 3 in front of the chlorine molecule

C. Leave the equation alone as it is already balanced

D. Place the coefficient 4 in front of carbon disulfide (CS2)

E. Place the coefficient 2 in front of carbon tetrachloride (CCl4)

The answer is B. Place the coefficient 3 in front of the chlorine molecule

Explanation:

You know the following equation:

CS₂ + Cl₂ ⇒ CCl₄ + S₂Cl₂

You must first look at the subscripts next to each atom to find the number of each atom in the equation, both in the reagents and in the products. If the element appears in more than one molecule, its amounts are added.

Reagents:

C: 1 atom S: 2 atoms Cl: 2 atoms

Products:

C: 1 atom S: 2 atoms Cl: 6 atoms (4 atoms in CCl₄ and 2 atoms in S₂Cl₂)

You can see that the only unbalanced element is chlorine, that is, it appears in different quantities in the reagent and in the product. Since chlorine appears as a diatomic molecule in reagents, you balance it by adding a coefficient 3 in front of this molecule, because, in a chemical equation you should never alter the subscripts. The coefficients can be multiplied by each subscript to obtain the number of atoms of each element.

In this case, if you multiply the coefficient 3 by the subscript 2 of the chlorine you get 6 atoms of Cl in the reagents.

So now you have:

In reagents:

C: 1 atom S: 2 atoms Cl: 6 atoms

In products:

C: 1 atom S: 2 atoms Cl: 6 atoms (4 atoms in CCl₄ and 2 atoms in S₂Cl₂)

Now you can see that there is the same amount of each element in the reagents and in the products. In other words, the reaction is balanced.

For that it was necessary place the coefficient 3 in front of the chlorine molecule

B. Place the coefficient 3 in front of the chlorine molecule

Explanation:

To balance the equation, placing 3 in front of the chlorine molecule makes it balanced.

Running a check of atoms on both left and right hand side of the equation shows that it obeys the law of conservation by mass.