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

1 Answer
Jun 12, 2017

P_(4(s)) + 10Cl_(2(g)) -> 4PCl_(5(g))

Explanation:

We'll balance the most complex compound first: PCl_5

The easiest way to do this is to balance the P since the ratio of P:PCl_5 in PCl_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 PCl_5 4 to do this.

P_(4(s)) + Cl_(2(g)) -> 4PCl_(5(g))

Now our phosphorus atoms are balanced. We see that in the reactants we have 2Cl atoms (Cl_2 being diatomic), and in the products we have 4 x 5=20 Cl atoms (4 molecules of PCl_5, 5 Cl atoms per molecule). By making the coefficient of Cl_2 in the reactants side 10, we end up with 20 Cl atoms on each side. Overall, we now have 4 P and 20 Cl on the reactants side, and 4P and 20Cl on the products side. The equation has been balanced.