Question

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

Answer 1

• Oxide: sodium oxide `"Na"_2 "O"`
• Peroxide: sodium peroxide `"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 `"Li"`, `"Na"`, and `"K"`.

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

`"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 `"BOH"` where `"B"` is of molar mass `x` would be `x + 17`.

One formula unit, or `x + 17 color(white)(l) "g"`, of the base would contain `16 color(white)(l) "g"` of oxygen atoms. `6.4 color(white)(l) "g"` of the base would thus contains `6.4 xx (16)/(x + 17) color(white)(l) "g"` of oxygen atoms.

The last process adds `200 - 6.4 = 193.6 color(white)(l) "g"` of water to the base. The `193.6 color(white)(l) "g"` of water `"H"_2"O"` would contain `193.6 * (16) /(18) = 172.09 color(white)(l) "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("oxygen") = m("oxygen in BOH") + m("oxygen in water")`
`6.4 xx (16)/(x + 17) + 172.09 = 174.648`

Solving the equation for `x` yields:

`x ~~ 23.01` which correspond to the molar mass of sodium `"Na"`, `22.99`. Therefore the oxide and peroxide are sodium oxide `"Na"_2"O"` and sodium peroxide `"Na"_2"O"_2`, respectively. Knowledge mass of the initial, anhydrous mixture (`5.6 color(white)(l) "g"`) is not likely required unless the question asks for the exact composition of that mixture.