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

1 Answer
Nov 30, 2015

H_3PO_4 + 5HCl rarr PCl_5 + 4H_2O

Explanation:

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

H_3PO_4 + HCl rarr PCl_5 + H_2O (unbalanced)

*left side: *
H = 3 + 1 (do not add this up yet)
P = 1
O = 4
Cl = 1

*right side: *
H = 2
P = 1
O = 1
Cl = 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_3PO_4 + color (red) 5HCl rarr PCl_5 + H_2O

left side: *
H = 3 + (1 x color (red) 5) = 8
P = 1
O = 4
Cl = (1 x color (red) 5) =
5*

right side: *
H = 2
P = 1
O = 1
Cl =
5*

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

Step 3: Continue balancing atoms.

H_3PO_4 + 5HCl rarr PCl_5 + color (blue) 4H_2O

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

right side: *
H = (2 x color (blue) 4) =
8
P = 1
O = (1 x color (blue) 4) =
4
Cl =
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_3PO_4 + 5HCl rarr PCl_5 + 4H_2O

The equation is now balance.