Question

a. write the balanced equation, including phase symbols, for the reaction between hi and manganese (iii) hydroxide.

mole map
throughout this homework, you may find it useful to refer to the mole - map below to remember what conversion factor(s) to use.
what we know about compound a
what we want to know about compound b

Explanation:

Step1: Identify Reactants and Products

The reactants are hydroiodic acid ($HI$) (aqueous, since it's an acid in solution) and manganese(III) hydroxide ($Mn(OH)_3$) (solid, as hydroxides of transition metals are typically insoluble). The reaction is an acid - base (neutralization) reaction, so the products will be a salt (manganese(III) iodide, $MnI_3$) and water ($H_2O$).

Step2: Write Unbalanced Equation

The unbalanced equation is: $HI(aq)+Mn(OH)_3(s)
ightarrow MnI_3(aq)+H_2O(l)$

Step3: Balance Iodine Atoms

On the left side, we have 1 $I$ atom (from $HI$), and on the right side, we have 3 $I$ atoms (from $MnI_3$). So we put a coefficient of 3 in front of $HI$: $3HI(aq)+Mn(OH)_3(s)
ightarrow MnI_3(aq)+H_2O(l)$

Step4: Balance Hydrogen and Oxygen Atoms

Now, let's balance the $H$ and $O$ atoms. On the left side, from $3HI$ we have 3 $H$ atoms, and from $Mn(OH)_3$ we have 3 $H$ atoms (from the 3 $OH^-$ groups) and 3 $O$ atoms. So total $H$ atoms on the left: $3 + 3=6$, and $O$ atoms: 3. On the right side, from $H_2O$ we have 2 $H$ atoms per molecule and 1 $O$ atom per molecule. Let's let the coefficient of $H_2O$ be $x$. Then for $H$: $2x = 6$ (so $x = 3$), and for $O$: $x=3$. So we put a coefficient of 3 in front of $H_2O$: $3HI(aq)+Mn(OH)_3(s)
ightarrow MnI_3(aq)+3H_2O(l)$

Answer:

$3HI(aq) + Mn(OH)_3(s)
ightarrow MnI_3(aq) + 3H_2O(l)$