Question

How do you balance `H_2SO_4 + B(OH)_3 -> B_2(SO_4)_3 + H_2O`?

Answer 1

`3H_2SO_4 + 2B(OH)_3 → B_2(SO_4)_3 + 6H_2O`

Explanation:

`H_2SO_4 + B(OH)_3 → B_2(SO_4)_3 + H_2O`

We have:
LHS
H= 2 + 3= 5
S= 1
B= 1
O= 4 + 3= 7

RHS
H= 2
S= 3
B=2
O= 7 + 1= 8

We balance the equation by making the number of each element the same on both sides.

Starting with the Hydrogen:
`3H_2SO_4 + B(OH)_3 → B_2(SO_4)_3 + 6H_2O`

LHS
H= 6 (from `H_2SO_4`) + 6 (from `(OH)_3`)= 12

RHS
H= 12 (from `H_2O`)

Then the Sulphur:
`3H_2SO_4 + B(OH)_3 → B_2(SO_4)_3 + 6H_2O`

LHS
S= 3 (from adding the 3 in front of the `H_2SO_4`)

RHS
S= 3

Then the Boron:
`3H_2SO_4 + 2B(OH)_3 → B_2(SO_4)_3 + 6H_2O`

Add a 2 in front of `B(OH)_3` so that

LHS: B= 2 and RHS: B=2

Finally, the Oxygen:
`3H_2SO_4 + 2B(OH)_3 → B_2(SO_4)_3 + 6H_2O`

All we have to do here is make sure everything is adding up to the same number on both sides.

LHS
O= 7 (from `3H_2SO_4`) + 6 (from `2B(OH)_3`) = 13

RHS
O= 7 (from `B_2(SO_4)_3`) + 6 (from `6H_2O`) = 13

THE END whew!