Balance each of the following neutralization equations:

a.${\mathrm{HNO}}_3\left(aq\right)+\mathrm{Ba}\left(\mathrm{OH}{)}_2\right(s)⟶{\mathrm{H}}_2\mathrm{O}(l)+\mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2(aq)$

b.localid="1654005825158" ${\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+\mathrm{Al}\left(\mathrm{OH}{)}_3\right(s)⟶{\mathrm{H}}_2\mathrm{O}(l)+{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3(aq)$

A neutralization is a chemical reaction in which acids and bases are quantitatively combined to form salts and water. It is a powerful neutralizing methods that combines manner,$H+$ions and $OH-$ions to form water.

a.Step 1 -Records by Reactant and Products

${\mathrm{HNO}}_3\left(aq\right)+\mathrm{Ba}\left(\mathrm{OH}{)}_2\right(\mathrm{S})\to \mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2(aq)+{\mathrm{H}}_2\mathrm{O}(l)$

Step 2- Balance ${\mathrm{H}}^{+}$in acid and base. Placing $2$ahead of ${\mathrm{HNO}}_3$makes $2{\mathrm{H}}^{+}$from $2\mathrm{OH}¯$in $\mathrm{Ba}(OH{)}_2$.

$2{\mathrm{HNO}}_3\left(aq\right)+\mathrm{Ba}\left(\mathrm{OH}{)}_2\right(\mathrm{S})\to \mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2(aq)+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Step 3- Balance ${\mathrm{H}}_2\mathrm{O}$with ${\mathrm{H}}^{+}$

$2{\mathrm{HNO}}_3\left(aq\right)+\mathrm{Ba}\left(\mathrm{OH}{)}_2\right(\mathrm{S})\to \mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2(aq)+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Step 4- To the acid and base, add the salt of the remaining ion.

$2{\mathrm{HNO}}_3\left(aq\right)+\mathrm{Ba}\left(\mathrm{OH}{)}_2\right(\mathrm{S})\to \mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2(aq)+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

b.Step 1 - Records by Reactant and merchandise

${\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+\mathrm{Al}\left(\mathrm{OH}{)}_3\right(aq)\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3(aq)+{\mathrm{H}}_2\mathrm{O}(l)$

Step 2 -Balance the ${\mathrm{H}}^{+}$within the acid and within the base$3$before of ${\mathrm{H}}_2{\mathrm{SO}}_4$and $2$before of $\mathrm{Al}(\mathrm{OH}{)}_2$provides $6{\mathrm{H}}^{+}$for the $6{\mathrm{OH}}^{-}$in $Al\left(OH\right)₃.$

$3{\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+2\mathrm{Al}\left(\mathrm{OH}{)}_3\right(aq)\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3(aq)+{\mathrm{H}}_2\mathrm{O}(l)$

Step 3-Balance the ${\mathrm{H}}_2\mathrm{O}$with ${\mathrm{H}}^{+}$and

$3{\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+2\mathrm{Al}\left(\mathrm{OH}{)}_3\right(aq)\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3(aq)+6{\mathrm{H}}_2\mathrm{O}(l)$

Step 4 -Within the acid and base, write the salt from the leftover ion.

$3{\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+2\mathrm{Al}\left(\mathrm{OH}{)}_3\right(aq)\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3(aq)+6{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$