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

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Answer 1 · 2017-05-10T20:24:08

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

#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!