# How do you balance H_3PO_4 + HCI -> PCl_5 + H_2O?

Nov 30, 2015

${H}_{3} P {O}_{4}$ + $5 H C l$ $\rightarrow$ $P C {l}_{5}$ + $4 {H}_{2} O$

#### Explanation:

Step 1: Tally all the atoms based on the subscripts.

${H}_{3} P {O}_{4}$ + $H C l$ $\rightarrow$ $P C {l}_{5}$ + ${H}_{2} O$ (unbalanced)

left side:
$H$ = 3 + 1 (do not add this up yet)
$P$ = 1
$O$ = 4
$C l$ = 1

right side:
$H$ = 2
$P$ = 1
$O$ = 1
$C l$ = 5

Step 2: Find the easiest atom to balance. Remember, in balancing you CANNOT change the subscripts - what you CAN do is multiply the subscripts by an appropriate coefficient. Also remember that with bonded substances, you have to multiply the coefficient, not only to the atom that you intend to balance, but also to whatever other atoms that it is bonded with.

${H}_{3} P {O}_{4}$ + $\textcolor{red}{5} H C l$ $\rightarrow$ $P C {l}_{5}$ + ${H}_{2} O$

left side:
$H$ = 3 + (1 x $\textcolor{red}{5}$) = 8
$P$ = 1
$O$ = 4
$C l$ = (1 x $\textcolor{red}{5}$) = 5

right side:
$H$ = 2
$P$ = 1
$O$ = 1
$C l$ = 5

But since $H C l$ is a substance, the coefficient 5 should also be applied to the $H$ atom.

Step 3: Continue balancing atoms.

${H}_{3} P {O}_{4}$ + $5 H C l$ $\rightarrow$ $P C {l}_{5}$ + $\textcolor{b l u e}{4} {H}_{2} O$

left side:
$H$ = 3 + (1 x 5) = 8
$P$ = 1
$O$ = 4
$C l$ = (1 x 5) = 5

right side:
$H$ = (2 x $\textcolor{b l u e}{4}$) = 8
$P$ = 1
$O$ = (1 x $\textcolor{b l u e}{4}$) = 4
$C l$ = 5

Step 3: Check if the total numbers per atom on the left is the same with the total numbers per atom on the side of the equation.

${H}_{3} P {O}_{4}$ + $5 H C l$ $\rightarrow$ $P C {l}_{5}$ + $4 {H}_{2} O$

The equation is now balance.