Question

Sodium superoxide${\text{Na}}{{\text{O}}_{\text{2}}}$ can’t be prepared directly from sodium, but form ${\text{N}}{{\text{a}}_2}{{\text{O}}_{\text{2}}}$, Then the amount of ${\text{N}}{{\text{a}}_2}{{\text{O}}_{\text{2}}}$ required to prepare $11$ g of ${\text{Na}}{{\text{O}}_{\text{2}}}$ is:
A. $7.2$ g
B. $7.8$ g
C. $3.6$ g
D. $3.9$ g

Answer 1

Hint:To determine the number of moles and grams of any reactant or product balanced equation is required. After writing the balanced equation, by comparing the number of moles of reactant and product, the amount of sodium peroxide can be determined. We can determine the number of gram weights by using the mole formula.

Formula used: ${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$

Complete answer:
${\text{Na}}{{\text{O}}_{\text{2}}}$ can be prepared from ${\text{N}}{{\text{a}}_2}{{\text{O}}_{\text{2}}}$.
The reaction for the formation of sodium oxide form sodium peroxide is as follows:
${\text{N}}{{\text{a}}_2}{{\text{O}}_{\text{2}}}\, + \,{{\text{O}}_2}\, \to \,2{\text{Na}}{{\text{O}}_2}$
Determine the mole of sodium oxide ${\text{Na}}{{\text{O}}_{\text{2}}}$ as follows;
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
Molar mass of sodium oxide ${\text{Na}}{{\text{O}}_{\text{2}}}$ is $55\,{\text{g/mol}}$
Substitute $55\,{\text{g/mol}}$ for molar mass and $11$ g for mass of ${\text{Na}}{{\text{O}}_{\text{2}}}$.
$\Rightarrow {\text{Mole}}\,{\text{ = }}\,\dfrac{{11\,{\text{g}}}}{{{\text{55}}\,{\text{g/mol}}}}$
$\Rightarrow {\text{Mole}}\,{\text{ = }}\,0.2$
According to the balanced equation, one-mole sodium peroxide is giving two moles of sodium oxide so, the mole of sodium peroxide, required to obtained $0.2$ mole of sodium oxide is,
$2\,{\text{mol}}\,\,{\text{Na}}{{\text{O}}_{\text{2}}}\, = \,1\,\,{\text{mol}}\,\,{\text{N}}{{\text{a}}_2}{{\text{O}}_2}$
$\Rightarrow 0.2\,{\text{mol}}\,\,{\text{Na}}{{\text{O}}_{\text{2}}}\, = \,0.1\,\,{\text{mol}}\,\,{\text{N}}{{\text{a}}_2}{{\text{O}}_2}$
So, $0.1\,$ mole of sodium peroxide is required to prepare $11$ g of ${\text{Na}}{{\text{O}}_{\text{2}}}$.
Use mole formula to determine the gram amount of sodium peroxide ${\text{N}}{{\text{a}}_2}{{\text{O}}_2}$ as follows:
Molar mass of sodium peroxide ${\text{N}}{{\text{a}}_2}{{\text{O}}_2}$ is $78\,{\text{g/mol}}$
Substitute $78\,{\text{g/mol}}$ for molar mass and $0.1$ mole for mass of ${\text{N}}{{\text{a}}_2}{{\text{O}}_2}$.
$\Rightarrow {\text{0}}{\text{.1}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{78\,{\text{g/mol}}}}$
$\Rightarrow \,{\text{mass}}\,{\text{ = }}\,{\text{0}}{\text{.1}}\,\, \times 78\,{\text{g/mol}}$
$\Rightarrow \,{\text{mass}}\,{\text{ = }}\,7.8\,{\text{g}}$
So, $7.8\,{\text{g}}$ of sodium peroxide required to prepare $11$ g of ${\text{Na}}{{\text{O}}_{\text{2}}}$.

Therefore, option (B) $7.8\,{\text{g}}$ is correct.

Note:
In place of moles, we can compare the gram weight also. The gram weight of sodium peroxide ${\text{N}}{{\text{a}}_2}{{\text{O}}_{\text{2}}}$ is $78$ and the gram weight of two moles of sodium oxide ${\text{Na}}{{\text{O}}_{\text{2}}}$ is $110$. According to balanced equation, $110$ gram of sodium oxide is obtaining from $78$ g of sodium peroxide so, $11$ g will be obtained from,
$110\,{\text{g}}\,\,{\text{Na}}{{\text{O}}_{\text{2}}}\, = \,78\,{\text{g}}\,\,{\text{N}}{{\text{a}}_2}{{\text{O}}_2}$
$11\,{\text{g}}\,\,{\text{Na}}{{\text{O}}_{\text{2}}}\, = \,7.8\,{\text{g}}\,\,{\text{N}}{{\text{a}}_2}{{\text{O}}_2}$
Stoichiometry measurements are used to determine the amount of reactant or product from the given amounts. Stoichiometry measurements give quantitative relations among the amounts of various species of a reaction. To determine the stoichiometry relations a balanced equation is necessary.