How do you balance $HBr + Mg(OH)_2 -> MgBr_2 + H_2O$ ?

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Answer 1 · 2016-04-01T14:50:19

i will tell u the steps.

Let the equation be

a $HBr$ + b $Mg(OH)_2$ $->$ c $MgBr_2+d H_2O$

Equating $H$ -atoms on both sides,

a+2b=2d

Equating $Br$ on both sides,

a=2c

Equating $Mg$ on both sides,

b=c

Equating $O$ on both sides,

2b=d

So..........a+2b=2d
$rArr$ 2c+2(c)=2d
$rArr$ 4c=2d
$rArr$ d=2c

Finally, substituting a,b,c,d

2c $HBr$ + c $Mg(OH)_2$ $->$ c $MgBr_2$ + 2c $H_2O$

2 $HBr$ + $Mg(OH)_2$ $->$ $MgBr_2$ + 2 $H_2O$

How do you balance HBr + Mg(OH)_2 -> MgBr_2 + H_2OHBr + Mg(OH)_2 -> MgBr_2 + H_2O?