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]
