The only thing that caused the mass to decrease is the reaction of #Fe_2O_3#, therefore the #SiO_2# did not react. Thus allows us to say that the #1.5-1.46=color(purple)(0.04# gram difference is caused by the #O_2#.
This is because the #Fe_3O_4# is also solid and will stay be measured while weighing, but the #O_2# will not be measured and will cause the difference. Good, now we know that 0.04 gram #O_2# is produced!
We use this to calculate how much #Fe_2O_3 # has reacted. We use the chemical equation that you provided.
#Fe_"2"O_"3"(s) -> Fe_"3"O_"4"(s)+O_"2"(g)#
We first have to balance this reaction, since it is not allowed by the Law of conservation of mass (you can see that there are 2 #Fe# on the left side and 3 #Fe# on the right side of the arrow). The easiest way to balance the equation is by first correcting for the #Fe#.
#3Fe_"2"O_"3"(s) -> 2Fe_"3"O_"4"(s)+O_"2"(g)#
Now on both sides, we have 6 #Fe#, but the #O#-atoms aren't right yet. We count #3xx3=9# O-atoms on the left and #2xx4=8# O-atoms on the right. We cannot have a #1/2#, therefore we double all the other numbers! We obtain:
#color(red)(6Fe_"2"O_"3"(s) -> 4Fe_"3"O_"4"(s)+O_"2"(g)#
Find the molar values of the compound above or calculate them with the masses of the atoms:
#Fe_"2"O_"3"#=159.69 #"gram"xx"mol"^(-1)#
#O_2#=32.00 #"gram"xx"mol"^(-1)#
#mol=("mass in " color(blue)( gram))/("molar mass in" color(blue)(("gram")/("mol)))#
#(0.04 color(red)(cancel(color(blue)(gram))))/(32.00 color(blue)(color(red)(cancel(color(blue)("gram")))/"mol"))=0.00125 color(blue)(" mol")#
We make a nice scheme to do the other calculations
#color(white)(aaa) color(red)(6Fe_"2"O_"3"(s) -> 4Fe_"3"O_"4"(s)+O_"2"(g)) color(white)(aaa)color(orange)("reaction")#
#color(white)(aaaa)6color(white)(aaaaaaaa):color(white)(aaaaa)4color(white)(aaaa):color(white)(aaa)1color(white)(aaaa)color(orange)("mol ratio")#
#color(white)(aaaa)"X"color(white)(aaaaaaaaaaaaaaaaaaa)0.00125color(white)(aaaa)color(orange)("mol")#
We use the mol ratio to calculate the amount of mol #Fe_2O_3#
#"X"=(0.00125 color(blue)(" mol ")xx6)/1=0.0075 color(blue)(" mol " Fe_2O_3)#
From this we calculate the mass of #Fe_2O_3#
#0.0075 xx159.69=1.1976 color(blue)(" gram " Fe_2O_3)#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#
Now we use this formula:
To calculate the mass percentage of #Fe_2O_3# in the sample, we have to use:
#color(green)(("Mass " Fe_2O_3)/("Total mass of sample")xx100%="Mass percentage of " Fe_2O_3#
#1.1976/1.5xx100%=79.8%#
Therefore the answer is A.
