# How many grams of potassium nitrate, KNO_3. are formed when 102 grams of nitric acid, HNO_3, react with potassium hydroxide KOH?

May 22, 2017

$164 g K N {O}_{3}$

#### Explanation:

Let's start by creating the chemical equation for this reaction:

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

This is a neutralization reaction, in which an acid ($H N {O}_{3}$) and a base ($K O H$) react to form a salt and (most often liquid) water.

We'll go about this problem by first using the molar mass of $H N {O}_{3}$ to calculate the number of moles present, then use the stoichiometric relationships (the coefficients, which are all $1$ in this case), to calculate the moles of $K N {O}_{3}$ that form, and lastly use its molar mass to find how many grams were formed.

Using dimensional analysis, the whole procedure is

102cancel(gHNO_3)((1cancel(molHNO_3))/(63.02cancel(gHNO_3)))((1cancel(molKNO_3))/(1cancel(molHNO_3)))((101.11gKNO_3)/(1cancel(molKNO_3))) = color(red)(164gKNO_3

May 22, 2017

There would be $\text{164 g KNO"_3}$.

#### Explanation:

Balanced Equation

$\text{HNO"_3("aq") + "KOH(aq)}$$\rightarrow$$\text{KNO"_3("aq") + "H"_2"O(l)}$

This is a neutralization reaction, which is a type of double replacement reaction. $\text{KNO"_3}$ is in aqueous solution, which means the $\text{K"^+}$ and ${\text{NO"_3}}^{-}$ ions are dissociated and there is no discrete compound of $\text{KNO"_3}$ as a product. However, if the reaction went to completion, and the only products were potassium nitrate and water, you could evaporate the water to isolate the potassium nitrate, and you should get the calculated mass $\pm$. Regardless, the purpose of this question is to help you learn how to do stoichiometry problems, so I will go ahead and answer as if $\text{KNO"_3}$ were a discrete product, which it would be if the water were evaporated.

The process goes like this:

color(red)("given mass HNO"_3"$\rightarrow$color(red)("moles HNO"_3"$\rightarrow$color(red)("moles KNO"_3"$\rightarrow$color(red)("mass KNO"_3"

We will need the MOLAR MASSES of $\text{HNO"_3}$ and $\text{KNO"_3}$.
Multiply the molar mass of each element by its subscript and add. The molar mass is the element's atomic weight on the periodic table in g/mol.

${\text{HNO}}_{3} :$(1xx1.008"g/mol H")+(1xx14.007"g/mol N")+(3xx15.999"g/mol O")="63.012 g/mol HNO"_3"

${\text{KNO}}_{3} :$(1xx39.0983"g/mol K")+(1xx14.007"g/mol N")+(3xx15.999"g/mol O")="101.102 g/mol KNO"_3"

color(blue)("Given Mass of Nitric Acid to Moles"
Multiply the given mass of $\text{HNO"_3}$ by the inverse of its molar mass.

102color(red)cancel(color(black)("g HNO"_3))xx(1"mol HNO"_3)/(63.012color(red)cancel(color(black)("g HNO"_3)))="1.6187 mol HNO"_3"

color(blue)("Moles Nitric Acid to Moles Potassium Nitrate"
Multiply the moles ${\text{HNO}}_{3}$ by the mole ratio between ${\text{HNO}}_{3}$ and ${\text{KNO}}_{3}$ so that mol ${\text{HNO}}_{3}$ cancels.

1.6187color(red)cancel(color(black)("mol HNO"_3))xx(1"mol KNO"_3)/(1color(red)cancel(color(black)("mol HNO"_3)))="1.6187 mol KNO"_3"

color(blue)("Moles Potassium Nitrate to Mass Potassium Nitrate"
Multiply mol $\text{KNO"_3}$ by its molar mass.

1.6187color(red)cancel(color(black)("mol KNO"_3))xx(101.102"g KNO"_3)/(1color(red)cancel(color(black)("mol KNO"_3)))="164 g KNO"_3 (rounded to three significant figures due to 102 g)