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. (?)

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.