# What is the product of this chemical equation?

## Two solutions, $N {a}_{2} C {O}_{3}$ and $H C l$, that were added together to form a product.

Aug 9, 2016

Here's what I got.

#### Explanation:

Depending on the mole ratio used when mixing those two solutions, sodium carbonate, ${\text{Na"_2"CO}}_{3}$, will react with hydrochloric acid, $\text{HCl}$, to produce

• aqueous sodium chloride, $\text{NaCl}$, and sodium bicarbonate, ${\text{NaHCO}}_{3}$
• aqueous sodium chloride and carbonic acid, ${\text{H"_2"CO}}_{3}$, which decomposes to give water and carbon dioxide, ${\text{CO}}_{2}$

When the two reactants are mixed in a $1 : 1$ mole ratio, you get

${\text{Na"_ 2"CO"_ (3(aq)) + "HCl"_ ((aq)) -> "NaCl"_ ((aq)) + "NaHCO}}_{3 \left(a q\right)}$

When you have equal numbers of moles of sodium carbonate and hydrochloric acid, all the moles of sodium carbonate are converted to moles of sodium bicarbonate.

When the two reactants are mixed in a $1 : 2$ mole ratio, you get

"Na"_ 2"CO"_ (3(aq)) + 2"HCl"_ ((aq)) -> 2"NaCl"_ ((aq)) + overbrace("H"_ 2"CO"_ (3(aq)))^(color(blue)("H"_ 2"O"_ ((l)) + "CO"_ (2(g))))

As the carbonic acid decomposes, you get

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

As the reaction proceeds, you should see carbon dioxide bubble out of solution.

The idea here is that once you get past the $1 : 1$ mole ratio for sodium carbonate and hydrochloric acid, the reaction involves sodium bicarbonate and hydrochloric acid

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

You can thus say that you have

"Na"_ 2"CO"_ (3(aq)) + "HCl"_ ((aq)) -> "NaCl"_ ((aq)) + color(red)(cancel(color(black)("NaHCO"_ (3(aq))))

color(red)(cancel(color(black)("NaHCO"_ (3(aq))))) + "HCl"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l)) + "CO"_ (2(g))
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

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