Question

How many atoms of hydrogen are present in \[{\text{7}}{\text{.8g}}\] of \[{\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}\] ?

Answer 1

Hint: We have to know that the moles are the one of the main units in chemistry. The moles of the molecule depend on the mass of the molecule and molecular mass of the molecule. Chemical reactions are measured by moles only. The number of equivalents of the reactant also depends on the moles of the molecule. The number of moles of the reactant and product are equal in the equilibrium reaction.
Formula used: We must have to know that the mole is defined as the given mass of the molecule is divided by the molecular mass of the molecule.
$Moles = \dfrac{{{\text{Mass of the molecule}}}}{{{\text{Molecular weight of the molecule}}}}$
The molecular weight of the molecule depends on the atomic weight of the atom present in the molecule. The molecular weight of the molecule is equal to the sum of the molecular weight of the atom and the number of the respective atoms in the molecule.
${\text{Molecular weight}} = {\text{Number of the atoms}} \times {\text{Atomic weight of the atom}}$
The number of atoms of the molecules is equal to the number of moles of the molecules multiplied by Avogadro’s number.
\[{\text{The Number Of Atoms = number of moles}} \times {\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\]

Complete answer:
Calculate the molecular mass of \[{\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}\] is
In aluminium hydroxide having one aluminium atom, three oxygen atoms and three hydrogen atoms.
The atomic weight of oxygen is \[{\text{16g}}\].
The atomic weight of hydrogen is \[{\text{1g}}\].
The atomic weight of aluminium is \[{\text{27g}}\].
The molecular mass of aluminium is,
\[{\text{Molecular}}\,{\text{weight}}\,{\text{ = }}\,{\text{Number}}\,{\text{of}}\,{\text{the}}\,{\text{atoms}}\, \times \,{\text{Atomic}}\,{\text{weight}}\,{\text{of}}\,{\text{the}}\,{\text{atom}}\]
Now we can substitute the known values we get,
\[ = 27 + 3 \times 16 + 3 \times 1\]
On simplification we get,
\[ = 78\]
The molecular mass of aluminium hydroxide is \[{\text{78g}}\].
The given mass of aluminium hydroxide is \[{\text{7}}{\text{.8g}}\].
Calculate the moles of given mass of aluminium hydroxide is,
$Moles = \dfrac{{{\text{Mass of the molecule}}}}{{{\text{Molecular weight of the molecule}}}}$
Now we can substitute the known values we get,
\[{\text{moles}}\,{\text{ = }}\dfrac{{{\text{7}}{\text{.8}}}}{{{\text{78}}}}\]
On simplification we get,
$Moles{\text{ = 0}}{\text{.1moles}}$
The number of hydrogen atoms in\[{\text{0}}{\text{.1moles}}\] of aluminium hydroxide,
\[{\text{The Number Of Atoms = number of moles}} \times {\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\]
\[{\text{The Number Of Hydrogen = 0}}{\text{.1}} \times {\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\]
\[ = 6.022 \times {\text{1}}{{\text{0}}^{{\text{22}}}}\]
According to the above calculation, there are \[6.022 \times {10^{22}}\] atoms of hydrogen are present in \[{\text{7}}{\text{.8g}}\] of \[{\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}\].

Note:
We must have to know that the number of moles of the reaction if calculated, in this way we can predict the reaction is not. The number of moles are used to find the yield percentage of the reaction. The moles are also used to find the formation of side products in the chemical reaction. The chemical formula of aluminium hydroxide is \[{\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}\]. The numerical value of Avogadro’s number is \[6.022 \times {10^{23}}\].