
How many moles of ${\text{KI}}$ are required to produce $0.4{\text{mol}}$ of ${{\text{K}}_2}{\text{Hg}}{{\text{I}}_4}$?
A. $0.4$
B. $0.8$
C. $3.2$
D. $1.6$
Answer
579.3k+ views
Hint: A mole is simply a unit of measurement. One mole of a molecule has $6.02 \times {10^{23}}$ atoms. This particular number is called Avogadro number. Relative atomic mass of an element in the periodic table is the mass of one mole of that atom. In the case of molecules, composition is given by its molecular formula.
Complete step by step answer:
Concentration is the amount of solute dissolved in a given amount of solution. There are different types of concentration units. Formula and molecular mass deal with individual atoms and molecules. Mole is the unit that relates the number of particles and mass.
The reaction between ${\text{KI}}$ and ${{\text{K}}_2}{\text{Hg}}{{\text{I}}_4}$ is given below:
${\text{HgC}}{{\text{l}}_2} + 4{\text{KI}} \to 2{\text{KCl}} + {{\text{K}}_2}{\text{Hg}}{{\text{I}}_4}$
From the above equation, it is obvious that $1{\text{mol}}$ of ${\text{HgC}}{{\text{l}}_2}$ reacts with $4{\text{mol}}$ of ${\text{KI}}$ to produce $2{\text{mol}}$ of ${\text{KCl}}$ and $1{\text{mol}}$ of ${{\text{K}}_2}{\text{Hg}}{{\text{I}}_4}$.
Or it can be said that $1{\text{mol}}$ of ${{\text{K}}_2}{\text{Hg}}{{\text{I}}_4}$ is produced from $4{\text{mol}}$ of ${\text{KI}}$.
So $0.4{\text{mol}}$ of ${{\text{K}}_2}{\text{Hg}}{{\text{I}}_4}$ is produced from $4 \times 0.4{\text{mol}}$ of ${\text{KI}}$.
So the number of moles of ${\text{KI}}$ $0.4 \times 4 = 1.6{\text{mol}}$ of ${\text{KI}}$.
Hence the option D is correct.
Additional information:
Molar mass is the collective name for atomic mass and molecular mass. We use moles to calculate the number of particles, molecules, concentrations, etc. It represents a very large quantity of items.
Note:
Moles relate the mass of a single atom in ${\text{amu}}$ to the mass in grams. Moles to atoms or molecules conversions can be done by multiplying with Avogadro number. Atoms or molecules to moles conversions can be done by dividing by Avogadro number. Moles to grams conversion is mole times molar mass.
Complete step by step answer:
Concentration is the amount of solute dissolved in a given amount of solution. There are different types of concentration units. Formula and molecular mass deal with individual atoms and molecules. Mole is the unit that relates the number of particles and mass.
The reaction between ${\text{KI}}$ and ${{\text{K}}_2}{\text{Hg}}{{\text{I}}_4}$ is given below:
${\text{HgC}}{{\text{l}}_2} + 4{\text{KI}} \to 2{\text{KCl}} + {{\text{K}}_2}{\text{Hg}}{{\text{I}}_4}$
From the above equation, it is obvious that $1{\text{mol}}$ of ${\text{HgC}}{{\text{l}}_2}$ reacts with $4{\text{mol}}$ of ${\text{KI}}$ to produce $2{\text{mol}}$ of ${\text{KCl}}$ and $1{\text{mol}}$ of ${{\text{K}}_2}{\text{Hg}}{{\text{I}}_4}$.
Or it can be said that $1{\text{mol}}$ of ${{\text{K}}_2}{\text{Hg}}{{\text{I}}_4}$ is produced from $4{\text{mol}}$ of ${\text{KI}}$.
So $0.4{\text{mol}}$ of ${{\text{K}}_2}{\text{Hg}}{{\text{I}}_4}$ is produced from $4 \times 0.4{\text{mol}}$ of ${\text{KI}}$.
So the number of moles of ${\text{KI}}$ $0.4 \times 4 = 1.6{\text{mol}}$ of ${\text{KI}}$.
Hence the option D is correct.
Additional information:
Molar mass is the collective name for atomic mass and molecular mass. We use moles to calculate the number of particles, molecules, concentrations, etc. It represents a very large quantity of items.
Note:
Moles relate the mass of a single atom in ${\text{amu}}$ to the mass in grams. Moles to atoms or molecules conversions can be done by multiplying with Avogadro number. Atoms or molecules to moles conversions can be done by dividing by Avogadro number. Moles to grams conversion is mole times molar mass.
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