Question

How many grams of NaOH is produced from $1.2\times {{10}^{2}}$ grams of $N{{a}_{2}}O$?
\[N{{a}_{2}}O+{{H}_{2}}O\to 2NaOH\]

Answer 1

Hint: This question can be solved using the mole ratio concept. Mole ratio uses coefficients in a balanced equation to determine the conversion factor that is used to relate the amount of moles of any two molecules in a chemical reaction.

Complete answer:
The mass of one mole of a substance is known as the molar mass of that substance. Its SI base unit is kg/mol but it is usually expressed in g/mol.
Now, the molecular mass of one mole of a compound is equal to the sum of atomic masses of all the elements present in the molecule.
We know that the atomic mass of sodium (${{M}_{Na}}$) is 22.99 g/mol,
The atomic mass of oxygen (${{M}_{O}}$) is 15.999 g/mol,
And the atomic mass of hydrogen (${{M}_{H}}$) is 1.008 g/mol.
So, the molecular mass or the mass of one mole of $N{{a}_{2}}O$ (\[{{M}_{N{{a}_{2}}O}}\]) is
\[\begin{align}
  & {{M}_{N{{a}_{2}}O}}=2\times {{M}_{Na}}+{{M}_{O}} \\
 & {{M}_{N{{a}_{2}}O}}=2\times 22.99+15.999 \\
 & {{M}_{N{{a}_{2}}O}}=61.979g/mol \\
\end{align}\]
It is given to us that $1.2\times {{10}^{2}}$ grams of $N{{a}_{2}}O$ is reacted.
Hence the number of moles of $N{{a}_{2}}O$ present in $1.2\times {{10}^{2}}$ grams of $N{{a}_{2}}O$ is
\[\begin{align}
  & {{n}_{N{{a}_{2}}O}}=\frac{{{w}_{N{{a}_{2}}O}}}{{{M}_{N{{a}_{2}}O}}} \\
 & {{n}_{N{{a}_{2}}O}}=\frac{1.2\times {{10}^{2}}}{61.979} \\
 & {{n}_{N{{a}_{2}}O}}\cong 1.94mol \\
\end{align}\]
From the balanced chemical reaction equation, we can see that 1 molecule of $N{{a}_{2}}O$ gives 2 molecules of NaOH.
Since different relative amounts of reactants and products can be considered by using the same proportionality, we can say that 1 mole of $N{{a}_{2}}O$ gives 2 moles of NaOH.
\[mole\text{ }ratio=\frac{{{n}_{N{{a}_{2}}O}}}{{{n}_{NaOH}}}=\frac{1}{2}\]
So, 1.94 moles of $N{{a}_{2}}O$ will give
$\begin{align}
  & {{n}_{NaOH}}=2\times {{n}_{N{{a}_{2}}O}} \\
 & {{n}_{NaOH}}=2\times 1.94 \\
 & {{n}_{NaOH}}=3.88mol \\
\end{align}$

Since the molecular mass or the mass of one mole of NaOH (\[{{M}_{NaOH}}\]) is
\[\begin{align}
  & {{M}_{NaOH}}={{M}_{Na}}+{{M}_{O}}+{{M}_{H}} \\
 & {{M}_{NaOH}}=22.99+15.999+1.008 \\
 & {{M}_{NaOH}}=39.997g/mol \\
\end{align}\]
Hence the mass of NaOH produced will be
\[\begin{align}
  & {{w}_{NaOH}}={{n}_{NaOH}}\times {{M}_{NaOH}} \\
 & {{w}_{NaOH}}=3.88\times 39.997 \\
 & {{w}_{NaOH}}\cong 155.19g \\
\end{align}\]
So, 155.19 grams of NaOH is produced from $1.2\times {{10}^{2}}$ grams of $N{{a}_{2}}O$.

Note:
It should be noted that if one mole of a substance is present, it has exactly the Avogadro number (${{N}_{A}}=6.022\times {{10}^{23}}$) of particles and the mass of one mole of a compound is equivalent to the sum of the mass of all the particles contained in one mole of a substance. This can also be used to define the molar mass of a substance.