# Question 44ec4

Oct 12, 2017

That's a silly way to say carbon dioxide!

$C {O}_{2} \left(g\right) + C a {\left(O H\right)}_{2} \left(a q\right) r i g h t \le f t h a r p \infty n s C {a}^{2 +} \left(a q\right) + {H}_{2} C {O}_{3} \left(a q\right)$

May be wrong, but it seems reasonable. The gas would interact with the hydroxide and water molecules to form carbonic acid.

There are also concentrations of $O {H}^{-}$ and ${H}^{+}$ leftover from this reaction and from the auto-ionization of water, but that's probably going too far into detail.

Oct 18, 2017

$\text{CO"_2"(g)" +"Ca(OH)"_2"(aq)" → "CaCO"_3"(s )" + "H"_2"O(l)}$

#### Explanation:

This reaction is a test for carbon dioxide.

You bubble the gas through lime water (an aqueous solution of calcium hydroxide).

If the gas is carbon dioxide, the solution will turn cloudy because of the formation of a precipitate of calcium carbonate.

You can think of the reaction as occurring in two steps:

1. Formation of carbonic acid

$\text{CO"_2"(g)" + "H"_2"O(l)" ⇌ "H"_2"CO"_3"(aq)}$

2. Neutralization of carbonic acid

$\text{Ca(OH)"_2"(aq)" + "H"_2"CO"_3"(aq)" → "CaCO"_3"(s )" + "2H"_2"O(l)}$

Add these two equations to get the overall equation.

"CO"_2"(g)" + color(red)(cancel(color(black)("H"_2"O(l)"))) ⇌ color(red)(cancel(color(black)("H"_2"CO"_3"(aq)")))
ul("Ca(OH)"_2"(aq)" + color(red)(cancel(color(black)("H"_2"CO"_3"(aq)"))) → "CaCO"_3"(s )" + color(red)(cancel(color(black)2))"H"_2"O(l)"color(white)(mmmmmmmmmmmmmmmmmml)#
$\text{Ca(OH)"_2"(aq)" + "CO"_2"(g)" ⇌ "CaCO"_3"(s )" + "H"_2"O(l)}$