Question

Balance each of the following neutralization equations:

a. $\mathrm{HCl}\left(aq\right)+\mathrm{Mg}\left(\mathrm{OH}{)}_2\right(s)⟶{\mathrm{H}}_2\mathrm{O}(l)+{\mathrm{MgCl}}_2(aq)$

b. localid="1654005351289" ${\mathrm{H}}_3{\mathrm{PO}}_4\left(aq\right)+\mathrm{LiOH}\left(aq\right)⟶{\mathrm{H}}_2\mathrm{O}\left(l\right)+{\mathrm{Li}}_3{\mathrm{PO}}_4\left(aq\right)$

Short answer

(Part a)$2\mathrm{HCl}\left(aq\right)+\mathrm{Mg}\left(\mathrm{OH}{)}_2\right(\mathrm{S})\to 2{\mathrm{H}}_2\mathrm{O}(\mathrm{I})+{\mathrm{MgCl}}_2(aq)$

(Part b)${\mathrm{H}}_3{\mathrm{PO}}_4\left(aq\right)+3\mathrm{LiOH}\left(aq\right)\to {\mathrm{Li}}_3{\mathrm{PO}}_4\left(aq\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Step-by-step solution

    Given Information (Part a)

(a) Make a list of the reactants and products.

$\mathrm{HCl}\left(aq\right)+\mathrm{Mg}\left(\mathrm{OH}{)}_2\right(\mathrm{S})\to {\mathrm{MgCl}}_2(aq)+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Given Explanation(Part a)

Step 2: Balance the ${\mathrm{H}}^{+}$in the acid with in the base.

$\mathrm{Mg}{\left(\mathrm{OH}\right)}_2$is replaced by $2{\mathrm{H}}^{+}$when a $2$is placed in front of$\mathrm{HCl}.$

$2\mathrm{HCl}\left(aq\right)+\mathrm{Mg}\left(\mathrm{OH}{)}_2\right(\mathrm{S})\to {\mathrm{MgCl}}_2(aq)+{\mathrm{H}}_2\mathrm{O}$

Balance the ${\mathrm{H}}_2\mathrm{O}$with ${\mathrm{H}}^{+}$and .

$2\mathrm{HCl}\left(aq\right)+\mathrm{Mg}\left(\mathrm{OH}{)}_2\right(\mathrm{S})\to {\mathrm{MgCl}}_2(aq)+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

In the acid and base, write the salt from the remaining ion.
$2\mathrm{HCl}\left(aq\right)+\mathrm{Mg}\left(\mathrm{OH}{)}_2\right(\mathrm{S})\to 2{\mathrm{H}}_2\mathrm{O}(l)+{\mathrm{MgCl}}_2(aq)$

    Given Information(Part b)

b) Make a list of the reactant and products.
${\mathrm{H}}_3{\mathrm{PO}}_4\left(aq\right)+\mathrm{LiOH}\left(aq\right)\to {\mathrm{Li}}_3{\mathrm{PO}}_4\left(aq\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Given Explanation(Part b)

b) Balance the ${\mathrm{H}}^{+}$in the acid with the $3{\mathrm{H}}^{+}$in $H_{\mathit{3}}P{O}_{\mathit{4}}$by putting a $3$in front of$\mathrm{LiOH}.$

${\mathrm{H}}_3{\mathrm{PO}}_4\left(aq\right)+3\mathrm{LiOH}\left(aq\right)\to {\mathrm{Li}}_3{\mathrm{PO}}_4\left(aq\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Balance the ${\mathrm{H}}_2\mathrm{O}$with ${\mathrm{H}}^{+}$and ${\mathrm{H}}^{+}$.

${\mathrm{H}}_3{\mathrm{PO}}_4\left(aq\right)+3\mathrm{LiOH}\left(aq\right)\to {\mathrm{Li}}_3{\mathrm{PO}}_4\left(aq\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

In the acid and base, write the salt from the remaining ion.

${\mathrm{H}}_3{\mathrm{PO}}_4\left(aq\right)+3{\mathrm{LiOH}}^{}\left(aq\right)\to {\mathrm{Li}}_3{\mathrm{PO}}_4\left(aq\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$