# Question f1eef

Aug 29, 2016

Here's what I got.

#### Explanation:

For part (A), calcium carbonate, ${\text{CaCO}}_{3}$, which is insoluble in water, will be dissolved in hydrochloric acid, $\text{HCl}$, to form aqueous calcium chloride, ${\text{CaCl}}_{2}$, and carbonic acid, ${\text{H"_2"CO}}_{3}$.

Because carbonic acid is highly unstable, it will decompose to produce water and release carbon dioxide, ${\text{CO}}_{2}$.

${\text{CaCO"_ (3(s)) + 2"HCl"_ ((aq)) -> "CaCl"_ (2(aq)) + "H"_ 2"O"_ ((l)) + "CO}}_{2 \left(g\right)} \uparrow$

Hydrochloric acid is a strong acid, which means that it dissociates completely in aqueous solution to release hydrogen cations, ${\text{H}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$.

${\text{CaCO"_ (3(s)) + 2 xx ["H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)] -> "Ca"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) + "CO}}_{2 \left(g\right)} \uparrow$

The complete ionic equation will look like this

${\text{CaCO"_ (3(s)) + 2"H"_ ((aq))^(+) + 2"Cl"_ ((aq))^(-) -> "Ca"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) + "CO}}_{2 \left(g\right)} \uparrow$

To get the net ionic equation, simply remove the spectator ions from both sides of the equation

${\text{CaCO"_ (3(s)) + 2"H"_ ((aq))^(+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) -> "Ca"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) + "H"_ 2"O"_ ((l)) + "CO}}_{2 \left(g\right)} \uparrow$

This will get you

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{CaCO"_ (3(s)) + 2"H"_ ((aq))^(+) -> "Ca"_ ((aq))^(2+) + "H"_ 2"O"_ ((l)) + "CO}}_{2 \left(g\right)} \uparrow} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In part (B), you're dealing with ammonium sulfate, ("NH"_4)_2"SO"_4, is a soluble ionic compound that dissociates completely in aqueous solution to form ammonium cations, ${\text{NH}}_{4}^{+}$, and sulfate anions, ${\text{SO}}_{4}^{2 -}$.

Sodium hydroxide, $\text{NaOH}$, is a strong base that dissociates completely to form sodium cations, ${\text{Na}}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$.

When these two solutions are mixed, a neutralization reaction will take place. This reaction will produce aqueous sodium sulfate, ${\text{Na"_2"SO}}_{4}$, ammonia, ${\text{NH}}_{3}$, and water.

("NH"_ 4)_ 2"SO"_ (4(aq)) + 2"NaOH"_ ((aq)) -> "Na"_ 2"SO"_ (4(aq)) + 2"NH"_ (3(aq)) + 2"H"_ 2"O"_ ((l))#

The complete ionic equation is

$2 {\text{NH"_ (4(aq))^(+) + "SO"_ (4(aq))^(2-) + 2"Na"_ ((aq))^(+) + 2"OH"_ ((aq))^(-) -> 2"Na"_ ((aq))^(+) + "SO"_ (4(aq))^(2-) + 2"NH"_ (3(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

Once again, eliminate the spectator ions

$2 {\text{NH"_ (4(aq))^(+) + color(red)(cancel(color(black)("SO"_ (4(aq))^(2-)))) + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + 2"OH"_ ((aq))^(-) -> color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)("SO"_ (4(aq))^(2-)))) + 2"NH"_ (3(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

to get the net ionic eqution

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{NH"_ (4(aq))^(+) + "OH"_ ((aq))^(-) -> "NH"_ (3(aq)) + "H"_ 2"O}}_{\left(l\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$