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 on the reactant side is equal to the number of atoms on the product side.
1. Identify the number of atoms of each element on both sides:
- Reactants:
- [tex]\(P_4(s)\)[/tex] has 4 phosphorus (P) atoms.
- [tex]\(Cl_2(g)\)[/tex] has 2 chlorine (Cl) atoms.
- Products:
- [tex]\(PCl_3(l)\)[/tex] has 1 phosphorus (P) atom and 3 chlorine (Cl) atoms.
2. Balance the phosphorus (P) atoms:
Since there are 4 phosphorus atoms in [tex]\(P_4\)[/tex] and 1 phosphorus atom in [tex]\(PCl_3\)[/tex], we need 4 molecules of [tex]\(PCl_3\)[/tex] to balance the phosphorus atoms.
[tex]\[ P_4(s) + Cl_2(g) \rightarrow 4PCl_3(l) \][/tex]
3. Balance the chlorine (Cl) atoms:
- On the products side, we now have [tex]\(4 \times 3 = 12\)[/tex] chlorine atoms because there are 4 molecules of [tex]\(PCl_3\)[/tex] with 3 chlorine atoms each.
- To balance 12 chlorine atoms on the reactant side, we need [tex]\( \frac{12}{2} = 6 \)[/tex] molecules of [tex]\(Cl_2\)[/tex].
4. Write the balanced equation:
[tex]\[ P_4(s) + 6Cl_2(g) \rightarrow 4PCl_3(l) \][/tex]
5. Check the number of atoms for each element to ensure they are balanced:
- Phosphorus: [tex]\(4 \text{ (left side)} = 4 \text{ (right side)}\)[/tex]
- Chlorine: [tex]\(12 \text{ (left side)} = 12 \text{ (right side)}\)[/tex]
The balanced equation is:
[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]\( \boxed{4} \)[/tex].
