# Question a1fd3

Aug 29, 2017

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

#### Explanation:

Calcium sulfite, ${\text{CaSO}}_{3}$, will react with hydrochloric acid, $\text{HCl}$, to produce aqueous calcium chloride, ${\text{CaCl}}_{2}$, and sulfurous acid, ${\text{H"_2"SO}}_{3}$.

However, sulfurous acid does not exist in aqueous solution as a molecule, it actually decomposes to form sulfur dioxide, ${\text{SO}}_{2}$, and water.

You can thus say that the balanced chemical equation that describes this reaction looks like this

"CaSO"_ (3(s)) + 2"HCl"_ ((aq)) -> "CaCl"_ (2(aq)) + color(red)(cancel(color(black)("H"_ 2"SO"_ (3(aq)))))

color(white)(aaaaaaaaa)color(red)(cancel(color(black)("H"_ 2"SO"_ (3(aq))))) -> "SO"_ (2(g)) + "H"_ 2"O"_ ((l))#
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

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

Now, to get the complete ionic equation, use the fact that hydrochloric acid is a strong acid and that calcium chloride is soluble in water.

${\text{CaSO"_ (3(s)) + overbrace(2"H"_ ((aq))^(+) + 2"Cl"_ ((aq))^(-))^(color(blue)(=2"HCl"_ ((aq)))) -> overbrace("Ca"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-))^(color(blue)(" = CaCl"_ (2(aq)))) + "SO"_ (2(g)) + "H"_ 2"O}}_{\left(l\right)}$

To find the net ionic equation, eliminate the spectator ions, i.e. the ions that are present on both sides of the chemical equation

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

You will end up with

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