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

Question

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

Wyatt Adkins · Accepted Answer

#H_3PO_4# + #5HCl# #rarr# #PCl_5# + #4H_2O#

Step 1: Using the subscripts as a guide, count every atom.

#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: Choose the atom that will be easiest to balance. Keep in mind that you cannot change the subscripts in balancing; all you can do is multiply the subscripts by the appropriate coefficient. Additionally, keep in mind that when dealing with bonded substances, you must multiply the coefficient not only with the atom you want to balance but also with any 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: Keep the atoms in balance.

#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: Verify that the total number of atoms on the left and the total number of atoms on the equation's side are the same.

#H_3PO_4# + #5HCl# #rarr# #PCl_5# + #4H_2O#

At this point, the equation is balanced.

Wyatt Atkins · Answer

undefined H₃PO₄ + 5 HCl → PCl₅ + 4 H₂O

How do you balance #H_3PO_4 + HCI -&gt; PCl_5 + H_2O#?

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