To balance the chemical equation:
[tex]\[ P_4(s) + Cl_2(g) \rightarrow PCl_3(l) \][/tex]
we need to ensure that the number of atoms of each element is the same on both sides of the equation.
1. Start by examining phosphorus (P). On the left side, we have [tex]\( P_4 \)[/tex], which contains 4 phosphorus atoms. Therefore, we need 4 phosphorus atoms on the right side as well.
2. Each [tex]\( PCl_3 \)[/tex] molecule contains 1 phosphorus atom. To get 4 phosphorus atoms, we need 4 [tex]\( PCl_3 \)[/tex] molecules. Thus, we place a coefficient of 4 in front of [tex]\( PCl_3 \)[/tex]:
[tex]\[ P_4(s) + Cl_2(g) \rightarrow 4PCl_3(l) \][/tex]
3. Next, examine chlorine (Cl). On the right side, with 4 [tex]\( PCl_3 \)[/tex] molecules, each containing 3 chlorine atoms, we have a total of:
[tex]\[ 4 \times 3 = 12 \text{ chlorine atoms} \][/tex]
4. On the left side, chlorine appears as [tex]\( Cl_2 \)[/tex] molecules, each containing 2 chlorine atoms. To get 12 chlorine atoms, we need:
[tex]\[ \frac{12}{2} = 6 \text{ } Cl_2 \][/tex]
So, we place a coefficient of 6 in front of [tex]\( Cl_2 \)[/tex]:
[tex]\[ P_4(s) + 6Cl_2(g) \rightarrow 4PCl_3(l) \][/tex]
The coefficient that should be placed in front of [tex]\( PCl_3 \)[/tex] to balance the equation is therefore:
[tex]\[ \boxed{4} \][/tex]
