# How do you balance P_4(s) + Cl_2(g) -> PCl_5(g)?

Jun 12, 2017

${P}_{4 \left(s\right)} + 10 C {l}_{2 \left(g\right)} \to 4 P C {l}_{5 \left(g\right)}$

#### Explanation:

We'll balance the most complex compound first: $P C {l}_{5}$

The easiest way to do this is to balance the $P$ since the ratio of $P : P C {l}_{5}$ in $P C {l}_{5}$ is $1 : 1$. On our reactants side, we have ${P}_{4}$ and thus we need a minimum of 4 $P$ on the products side to balance this. We'll make the coefficient of $P C {l}_{5}$ 4 to do this.

${P}_{4 \left(s\right)} + C {l}_{2 \left(g\right)} \to 4 P C {l}_{5 \left(g\right)}$

Now our phosphorus atoms are balanced. We see that in the reactants we have 2$C l$ atoms ($C {l}_{2}$ being diatomic), and in the products we have $4$ x $5 = 20 C l$ atoms (4 molecules of $P C {l}_{5}$, 5 $C l$ atoms per molecule). By making the coefficient of $C {l}_{2}$ in the reactants side 10, we end up with 20 $C l$ atoms on each side. Overall, we now have 4 $P$ and 20 $C l$ on the reactants side, and 4$P$ and 20$C l$ on the products side. The equation has been balanced.