This formula equation is unbalanced.

[tex]\[ P_4(s) + Cl_2(g) \rightarrow PCl_3(l) \][/tex]

Which coefficient should be placed in front of [tex]\[ PCl_3 \][/tex] to balance this equation?

A. 1
B. 2
C. 4
D. 6



Answer :

Sure, let's balance the chemical equation step-by-step.

First, we need to count the number of each type of atom on both sides of the equation.

Unbalanced Equation:
[tex]\[ P_4(s) + Cl_2(g) \rightarrow PCl_3(l) \][/tex]

Step 1: Count the atoms on each side

On the left side:
- [tex]\( P_4 \)[/tex]: 4 atoms of Phosphorus (P)
- [tex]\( Cl_2 \)[/tex]: 2 atoms of Chlorine (Cl)

On the right side:
- [tex]\( PCl_3 \)[/tex]: 1 atom of Phosphorus (P) and 3 atoms of Chlorine (Cl)

Step 2: Balance Phosphorus (P) atoms

Currently, we have 4 Phosphorus atoms on the left side and only 1 on the right side. To balance the Phosphorus atoms, we need 4 molecules of [tex]\( PCl_3 \)[/tex].

[tex]\[ P_4(s) + Cl_2(g) \rightarrow 4PCl_3(l) \][/tex]

Now we have 4 Phosphorus atoms on both sides of the equation.
- Left side: 4 Phosphorus (from [tex]\( P_4 \)[/tex])
- Right side: [tex]\( 4 \times 1 = 4 \)[/tex] Phosphorus (from [tex]\( 4PCl_3 \)[/tex])

Step 3: Balance Chlorine (Cl) atoms

Now let's check the Chlorine atoms. When we place the coefficient 4 in front of [tex]\( PCl_3 \)[/tex], we get:
[tex]\[ 4 \times 3 = 12 \][/tex] Chlorine atoms on the right side.

So, we need 12 Chlorine atoms on the left side. Since each molecule of [tex]\( Cl_2 \)[/tex] contains 2 Chlorine atoms, we need:
[tex]\[ \frac{12}{2} = 6 \][/tex] molecules of [tex]\( Cl_2 \)[/tex].

The balanced equation will be:
[tex]\[ P_4(s) + 6Cl_2(g) \rightarrow 4PCl_3(l) \][/tex]

Thus, the coefficient that should be placed in front of [tex]\( PCl_3 \)[/tex] to balance the equation is [tex]\( 4 \)[/tex].

Therefore, the correct answer is:
[tex]\[ 4 \][/tex]