# A mix of oxide and peroxide of the same monovalent metal has the mass of 5.6g.The mix is treated with an excess amount of water , forming 6.4g of base. (?)

## Over the obtained solution , water is added , which causes the formation of 200g of solution which contains 87.324% oxygen Identify the oxide and the peroxide

Jul 15, 2018
• Oxide: sodium oxide $\text{Na"_2 "O}$
• Peroxide: sodium peroxide ${\text{Na"_2 "O}}_{2}$

#### Explanation:

The question is asking for the identification of the cation of the oxide/peroxide mixture.

The question states that the metal is monovalent, or in other words, contain one single electron in its valence shell. It is thus an IUPAC group $1$ alkaline metal. Examples of alkaline metals include $\text{Li}$, $\text{Na}$, and $\text{K}$.

Oxides and peroxides of group one metals react with water to form the corresponding hydroxide (a.k.a. base). Let ${\text{B}}^{+}$ resembles the cation of an alkaline metal,

$\text{B"_2"O" + "H"_2"O" to 2color(white)(l) "BOH}$
2color(white)(l) "B"_2"O"_2 + 2 "H"_2"O" to 4 color(white)(l) "BOH" + "O"_2 (g)

$17$ is the molar mass for one formula unit of hydroxide ion. The formula mass of base $\text{BOH}$ where $\text{B}$ is of molar mass $x$ would be $x + 17$.

One formula unit, or $x + 17 \textcolor{w h i t e}{l} \text{g}$, of the base would contain $16 \textcolor{w h i t e}{l} \text{g}$ of oxygen atoms. $6.4 \textcolor{w h i t e}{l} \text{g}$ of the base would thus contains $6.4 \times \frac{16}{x + 17} \textcolor{w h i t e}{l} \text{g}$ of oxygen atoms.

The last process adds $200 - 6.4 = 193.6 \textcolor{w h i t e}{l} \text{g}$ of water to the base. The $193.6 \textcolor{w h i t e}{l} \text{g}$ of water $\text{H"_2"O}$ would contain $193.6 \cdot \frac{16}{18} = 172.09 \textcolor{w h i t e}{l} \text{g}$ of oxygen by mass. The 87.324 % * 200 color(white)(l) "g" = 174.648 color(white)(l) "g" of oxygen in the final base solution would attribute to the two sources. Therefore

$m \left(\text{oxygen") = m("oxygen in BOH") + m("oxygen in water}\right)$
$6.4 \times \frac{16}{x + 17} + 172.09 = 174.648$

Solving the equation for $x$ yields:

$x \approx 23.01$ which correspond to the molar mass of sodium $\text{Na}$, $22.99$. Therefore the oxide and peroxide are sodium oxide $\text{Na"_2"O}$ and sodium peroxide ${\text{Na"_2"O}}_{2}$, respectively. Knowledge mass of the initial, anhydrous mixture ($5.6 \textcolor{w h i t e}{l} \text{g}$) is not likely required unless the question asks for the exact composition of that mixture.