How would you complete and balance the acid base neutralizations reaction: $HBr + Ba(OH)_2 ->$ ?

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Answer 1 · 2017-03-27T02:58:42

$2"HBr" + "Ba(OH)"_2 -> "BaBr"_2 + 2 "H"_2"O"$

In a neutralization reaction, if the base is a hydroxide, then the products formed would be the salt and water.

acid + base $->$ salt + water

The water ( $"H"_2"O"$ ) is a result of the combination of the $"H"^+$ from the acid and the $"OH"^-$ from the base.

The salt ( $"BaBr"_2$ ) is simply the remaining anion of the acid ( $"Br"^-$ ) and the cation of the base ( $"Ba"^{2+}$ ).

The unbalanced equation is therefore

$"HBr" + "Ba(OH)"_2 -> "BaBr"_2 + "H"_2"O"$

To balance the equation, first notice that the number of Ba on both sides are the same. Ba is balanced.

Next, notice that the LHS is short of 1 x Br. So balance it by adding 1 unit of HBr.

$"2HBr" + "Ba(OH)"_2 -> "BaBr"_2 + "H"_2"O"$

Now, the RHS is missing 2 x H and 1 x O. Balance that by adding 1 units of $"H"_2"O"$ .

$"2HBr" + "Ba(OH)"_2 -> "BaBr"_2 + 2"H"_2"O"$

Now, all the elements have equal amounts on both sides of the equation. The chemical equation is now considered balanced.

How would you complete and balance the acid base neutralizations reaction: HBr + Ba(OH)_2 ->HBr + Ba(OH)_2 ->?