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

1 Answer
Jun 12, 2017

Answer:

#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 2#Cl# 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 4#P# and 20#Cl# on the products side. The equation has been balanced.