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

Md Ariful Islam

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Md Ariful Islam

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

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!

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