Question

classify the following reaction and balance the equation by entering the smallest possible integer coefficients. hi(aq) + ba(oh)₂(aq) → bai₂(aq) + h₂o(l)

Explanation:

Step1: Balance iodine atoms

There are 2 iodine atoms in $BaI_2$. So, we need 2 moles of $HI$ on the left - hand side to balance the iodine atoms.
$2HI(aq)+Ba(OH)_2(aq)\longrightarrow BaI_2(aq) + H_2O(l)$

Step2: Balance hydrogen and oxygen atoms

On the left - hand side, we have 4 hydrogen atoms and 2 oxygen atoms from $2HI$ and $Ba(OH)_2$. On the right - hand side, in $BaI_2$ there are no hydrogen and oxygen atoms. In $H_2O$, to balance the 4 hydrogen and 2 oxygen atoms, we need 2 moles of $H_2O$.
$2HI(aq)+Ba(OH)_2(aq)\longrightarrow BaI_2(aq)+2H_2O(l)$

This is an acid - base neutralization reaction where hydroiodic acid ($HI$) reacts with barium hydroxide ($Ba(OH)_2$) to form a salt ($BaI_2$) and water ($H_2O$).

Answer:

$2$; $1$; $1$; $2$