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

1 Answer
Jul 15, 2018
  • 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.