Posted by Deane on Thursday, November 15, 2012 at 7:42pm.
Trial and error.
HBr(aq)+Mg(OH)2(s)-->MgBr2(aq)+H2O(l)
Here is how I do these. It takes a little practice.
I look on the right and see Mg is ok but Br is Br2. So on the left I place a 2 for HBr.
2HBr(aq)+Mg(OH)2(s)-->MgBr2(aq)+H2O(l)
That takes care of everything except H and O. Since I know the 2 for HBr is right AND the 1 for Mg(OH)2 is right, I can count up H I have4 on the left and make the right equal to that. So I have 2H from 2HBr and I have 2H from (OH)2 which makes 4 H. On the right then I must place a 2 for 2H2O.
2HBr(aq)+Mg(OH)2(s)-->MgBr2(aq)+2H2O(l)
The only thing left is to check O and it SHOULD balance. On the left I have 2O atoms (from (OH)2) and on the right I have 2 O atoms (from 2H2O). Everything balances.
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