Question

What is the mass percent of the sample that was actually `"Fe"_2"O"_3`? The unbalanced reaction was `Fe_2O_3(s) -> Fe_3O_4(s) + O_2(g)`

Answer 1

The calculated mass percentage in the sample was 79.8%, which is closest to A.

Explanation:

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.