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

May 10, 2017

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 ${H}_{2} S {O}_{4}$) + 6 (from ${\left(O H\right)}_{3}$)= 12

RHS
H= 12 (from ${H}_{2} O$)

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 ${H}_{2} S {O}_{4}$)

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 $B {\left(O H\right)}_{3}$ 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 $3 {H}_{2} S {O}_{4}$) + 6 (from $2 B {\left(O H\right)}_{3}$) = 13

RHS
O= 7 (from ${B}_{2} {\left(S {O}_{4}\right)}_{3}$) + 6 (from $6 {H}_{2} O$) = 13

THE END whew!