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. (?)

Question

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

Natalie Anton · Accepted Answer

Oxide: sodium oxide #"Na"_2 "O"#
Peroxide: sodium peroxide #"Na"_2 "O"_2#

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.

Ellie Andrews · Answer

undefined To find the amount of oxide and peroxide in the mixture, we first need to determine the molar masses of the oxide and peroxide of the monovalent metal. Let's denote the mass of the oxide as $ x $ grams and the mass of the peroxide as $ 5.6 - x $ grams.

Next, we need to consider the chemical reactions involved. When the mixture is treated with water, it reacts to form a base. The reaction of the oxide with water yields one mole of base per mole of oxide, while the reaction of the peroxide with water yields two moles of base per mole of peroxide.

Using the molar masses of the oxide and peroxide, we can calculate the number of moles of each compound in the mixture. Then, we can determine the amount of base produced from each compound and sum them up to find the total amount of base produced.

Finally, we compare the calculated mass of base produced with the given mass of base (6.4g) to find the value of $ x $, which represents the mass of the oxide in the mixture. Once we have $ x $, we can find the mass of the peroxide as $ 5.6 - x $ grams.

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

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