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)

Jun 2, 2017

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 $F {e}_{2} {O}_{3}$, therefore the $S i {O}_{2}$ 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 $F {e}_{3} {O}_{4}$ 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 $F {e}_{2} {O}_{3}$ has reacted. We use the chemical equation that you provided.

$F {e}_{\text{2"O_"3"(s) -> Fe_"3"O_"4"(s)+O_"2}} \left(g\right)$
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 $F e$ on the left side and 3 $F e$ on the right side of the arrow). The easiest way to balance the equation is by first correcting for the $F e$.

$3 F {e}_{\text{2"O_"3"(s) -> 2Fe_"3"O_"4"(s)+O_"2}} \left(g\right)$
Now on both sides, we have 6 $F e$, but the $O$-atoms aren't right yet. We count $3 \times 3 = 9$ O-atoms on the left and $2 \times 4 = 8$ O-atoms on the right. We cannot have a $\frac{1}{2}$, 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:
$F {e}_{\text{2"O_"3}}$=159.69 ${\text{gram"xx"mol}}^{- 1}$
${O}_{2}$=32.00 ${\text{gram"xx"mol}}^{- 1}$

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

$\textcolor{w h i t e}{a a a} \textcolor{red}{6 F {e}_{\text{2"O_"3"(s) -> 4Fe_"3"O_"4"(s)+O_"2"(g)) color(white)(aaa)color(orange)("reaction}}}$
$\textcolor{w h i t e}{a a a a} 6 \textcolor{w h i t e}{a a a a a a a a} : \textcolor{w h i t e}{a a a a a} 4 \textcolor{w h i t e}{a a a a} : \textcolor{w h i t e}{a a a} 1 \textcolor{w h i t e}{a a a a} \textcolor{\mathmr{and} a n \ge}{\text{mol ratio}}$

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 $F {e}_{2} {O}_{3}$

"X"=(0.00125 color(blue)(" mol ")xx6)/1=0.0075 color(blue)(" mol " Fe_2O_3)
From this we calculate the mass of $F {e}_{2} {O}_{3}$
$0.0075 \times 159.69 = 1.1976 \textcolor{b l u e}{\text{ gram } F {e}_{2} {O}_{3}}$

$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
Now we use this formula:
To calculate the mass percentage of $F {e}_{2} {O}_{3}$ 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.