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

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

& {{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{aligned}$$ It is given to us that $1.2\times {{10}^{2}}$ grams of $N{{a}_{2}}O$ is reacted. Hence the number of moles of $N{{a}_{2}}O$ present in $1.2\times {{10}^{2}}$ grams of $N{{a}_{2}}O$ is $$\begin{aligned} & {{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{aligned}$$ From the balanced chemical reaction equation, we can see that 1 molecule of $N{{a}_{2}}O$ gives 2 molecules of NaOH. Since different relative amounts of reactants and products can be considered by using the same proportionality, we can say that 1 mole of $N{{a}_{2}}O$ gives 2 moles of NaOH. $$mole\text{ }ratio=\frac{{{n}_{N{{a}_{2}}O}}}{{{n}_{NaOH}}}=\frac{1}{2}$$ So, 1.94 moles of $N{{a}_{2}}O$ will give $\begin{aligned} & {{n}_{NaOH}}=2\times {{n}_{N{{a}_{2}}O}} \\\ & {{n}_{NaOH}}=2\times 1.94 \\\ & {{n}_{NaOH}}=3.88mol \\\ \end{aligned}$ Since the molecular mass or the mass of one mole of NaOH ($${{M}_{NaOH}}$$) is $$\begin{aligned} & {{M}_{NaOH}}={{M}_{Na}}+{{M}_{O}}+{{M}_{H}} \\\ & {{M}_{NaOH}}=22.99+15.999+1.008 \\\ & {{M}_{NaOH}}=39.997g/mol \\\ \end{aligned}$$ Hence the mass of NaOH produced will be $$\begin{aligned} & {{w}_{NaOH}}={{n}_{NaOH}}\times {{M}_{NaOH}} \\\ & {{w}_{NaOH}}=3.88\times 39.997 \\\ & {{w}_{NaOH}}\cong 155.19g \\\ \end{aligned}$$ 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.