Answer:
a: [tex]4KO_2(s) +2CO_2(g) = 2K_2CO_3(s) +3O_2(g)[/tex]
b: Oxygen is the only element which is getting oxidised as well reduced
c: [tex]mass of KO_2 =58.22 gram[/tex] and [tex]mass of O_2 =19.68 gram[/tex]
Explanation:
Part a: Balanced equation for the given reaction;
Rule:
step1 : first balance the metal
step2: Non metal except oxygen
step 3: lastly balance the oxygen
[tex]4KO_2(s) +2CO_2(g) = 2K_2CO_3(s) +3O_2(g)[/tex]
Part b: indicating oxidation number of each element both side
reactant side oxidation number:
K = +1
O= -0.5 and -2
C =+4
Product side oxidation number:
K = +1
O= 0 and -2
C =+4
From the above data it clearly that Oxygen is the only element which is getting oxidised as well reduced .
Part c: Mass calculation
from the balance equation it is clearly that 4 moles of [tex]KO_2[/tex] need 2 moles of [tex]CO_2[/tex] for complete reaction i.e. mole [tex]KO_2[/tex] is used double the mole of [tex]CO_2[/tex]
[tex]Mole of CO2 = 18/44 mol[/tex]
[tex]Mole of CO2 = 0.41 mol[/tex]
hence the mole of [tex]KO_2[/tex] is two times of [tex]CO_2[/tex]
[tex]Mole of KO_2 = 0.82 mol[/tex]
[tex]Mass of KO_2 = mole \times molecular weight[/tex]
[tex]mass of KO_2 =0.82 \times71 gram[/tex]
[tex]mass of KO_2 =58.22 gram[/tex]
From 2 mole of [tex]CO_2[/tex] 3 mole of [tex]O_2[/tex] is produced
From 1 mole of [tex]CO_2[/tex] 1.5 mole of [tex]O_2[/tex] is produced
From 0.41 mole of [tex]CO_2[/tex] [tex]1.5\times0.41[/tex] mole of [tex]O_2[/tex] is produced
[tex]Mass of O_2 = mole \times molecular weight[/tex]
[tex]mass of O_2 =1.5 \times0.41\times 32 gram[/tex]
[tex]mass of O_2 =19.68 gram[/tex]