Question

How many grams of potassium nitrate, $KN{O_3}$ are formed when 102 grams of nitric acid, $HN{O_3}$, reacts with potassium hydroxide KOH?

Answer 1

Hint: In this question, mass of nitric acid is given so the moles of nitric acid is calculated by dividing the mass of the compound by the molecular mass of the compound. The given reaction is a neutralization reaction.

Complete step by step answer:
The mass of nitric acid is 102 grams.
The reaction taking place between nitric acid and potassium hydroxide is shown below.
$HN{O_3} + KOH \to KN{O_3} + {H_2}O$
In this reaction, one mole of nitric acid reacts with one mole of potassium hydroxide to form one mole of potassium nitrate and one mole of water. This reaction is an example of neutralization reaction. In neutralization reaction, acid reacts with base to form salt and water.
The formula to calculate the number of moles is shown below.
$n = \dfrac{m}{M}$
Where,
n is the number of moles
m is the mass
M is the molar mass
The molar mass of nitric acid is 63 g/mol.
To calculate the moles of oxygen, substitute the values in the above equation.
$\Rightarrow n = \dfrac{{102}}{{63}}$
$\Rightarrow n = 1.619$mol
1.619 mol of nitric acid $HN{O_3}$ gives 1.619 mol of potassium nitrate $KN{O_3}$
The molecular mass of potassium nitrate is 101.1032 g/mol.
To calculate the mass of potassium nitrate, substitute the value of moles and molecular mass in the above equation.
$\Rightarrow 1.619 = \dfrac{m}{{101.1032}}$
$\Rightarrow m = 1.619 \times 101.1032$
$\Rightarrow m = 163.68$g $\approx 164$g
Therefore, 164grams of potassium nitrate, $KN{O_3}$ are formed when 102 grams of nitric acid, $HN{O_3}$, reacts with potassium hydroxide KOH.

Note:
The given reaction is already balanced, if the number of moles of compound reacting will be more than one then it should be multiplied with the calculated moles of the compound to get the final value of mole.