# How would you balance PCl5 + H2O -->H3PO4 + HCl?

Oct 25, 2015

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

#### Explanation:

Just create a tally sheet for all the elements involve.

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

Left side:
P = 1
Cl = 5
H = 2
O = 1

Right side:
P = 1
Cl = 1
H = 3 + 1 (DO NOT ADD IT UP YET)
O = 4

Start balancing the easiest element, in this case, the Cl.

Left side:
P = 1
Cl = 5
H = 2
O = 1

since $H C l$ is a substance, you need to also multiply its H by 5.

Right side:
P = 1
Cl = 1 x 5 = 5
H = 3 + (1 x 5) = 8
O = 4

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

Now the H atom on the left side is not balance with the right side, we have to multiply the H on the left side by 4.

Left side:
P = 1
Cl = 5
H = 2 x 4 = 8
O = 1 x 4 = 4

Again, since ${H}_{2} O$ is a substance, you also need to multiply its O by 4.

Right side:
P = 1
Cl = 1 x 5 = 5
H = 3 + (1 x 5) = 8
O = 4

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

The equation is now balanced.