# Question f8f5c

Dec 19, 2017

"74 g Ca"("OH")_2

#### Explanation:

Start by writing the balanced chemical equation that describes this neutralization reaction.

${\text{Ca"("OH")_ (2(aq)) + 2"HNO"_ (3(aq)) -> "Ca"("NO"_ 3)_ (2(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

Now, the trick here is to realize that in any chemical reaction, the total mass of the products must be equal to the total mass of the reactants $\to$ think the Law of Mass Conservation here.

So if you assume that $\text{126 g}$ of nitric acid took part in the reaction, then you can say that if the reaction produced

overbrace(" 36 g ")^(color(blue)("mass of water")) + overbrace(" 164 g ")^(color(blue)("mass of calcium nitrate")) = overbrace(" 200 g ")^(color(blue)("total mass of products"))

it must have consumed

overbrace(" 126 g ")^(color(blue)("mass of nitric acid")) + overbrace(" ? g ")^(color(blue)("mass of calcium hydroxide")) = overbrace(" 200 g ")^(color(blue)("total mass of reactants"))

This implies that the mass of calcium hydroxide that reacted is equal to

"mass Ca"("OH")_2 = "200 g " - " 126 g" = color(darkgreen)(ul(color(black)("74 g")))

To double-check the answer, use the molar masses of calcium nitrate, water, and nitric acid to convert all the masses to moles--I'll round all the values to the nearest whole number to make the calculations easier.

36 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18color(red)(cancel(color(black)("g")))) = "2 moles H"_2"O"

164 color(red)(cancel(color(black)("g"))) * ("1 mole Ca"("NO"_3)_2)/(164 color(red)(cancel(color(black)("g")))) = "1 mole Ca"("NO"_3)_2

126 color(red)(cancel(color(black)("g"))) * "1 mole HNO"_3/(126color(red)(cancel(color(black)("g")))) = "1 mole HNO"_3#

According to the balanced chemical equation, when $1$ mole of calcium hydroxide reacts with $2$ moles of nitric acid, the reaction produces $1$ mole of aqueous calcium nitrate and $2$ moles of water.

As you can see, the values that you have here are consistent with what you would need, i.e. $2$ moles of nitric acid, and what you would get, i.e. $1$ mole of calcium nitrate and $2$ moles of water, when $1$ mole of calcium hydroxide reacts.

Use the molar mass of calcium hydroxide to convert this to grams to get

$1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mole Ca"("OH")_2))) * "74 g"/(1color(red)(cancel(color(black)("mole Ca"("OH")_2)))) = color(darkgreen)(ul(color(black)("74 g}}}}$