# Question cd3fd

Dec 29, 2017

21%

#### Explanation:

Start by writing the balanced chemical equation that describes this reaction.

$3 {\text{Fe"_ ((s)) + 4"H"_ 2"O"_ ((g)) -> "Fe"_ 3"O"_ (4(s)) + 4"H}}_{2 \left(g\right)}$

Now, the trick here is to realize that after the reaction is complete, the total mass of the sample includes the mass of unreacted iron metal.

This means that you have

${\text{648 g" = "mass of unreacted Fe" + "mass of Fe"_3"O}}_{4}$

Let's take $x$ $\text{g}$ to be the mass of iron metal that reacted and $y$ $\text{g}$ the mass of iron(II, III) oxide that was produced by the reaction.

You can say that after the reaction is complete, you will have

648 color(red)(cancel(color(black)("g"))) = overbrace((600 -x)color(white)(.)color(red)(cancel(color(black)("g"))))^(color(blue)("mass of unreacted Fe")) + overbrace(" " y color(white)(.)color(red)(cancel(color(black)("g"))))^(color(blue)("mass of Fe"_3"O"_4))

which gets you

$648 = 600 - x + y \text{ " " } \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(1\right)}$

Now, you can use the molar masses of iron metal and iron(II, III) oxide, respectively, to express the mass of iron metal that reacted and the mass of iron(II, III) oxide that was produced in terms of moles.

$x \textcolor{w h i t e}{.} \text{g Fe" implies (x color(red)(cancel(color(black)("g"))))/(55.845color(red)(cancel(color(black)("g")))"mol"^(-1)) = (x/55.845)color(white)(.)"moles Fe}$

$y \textcolor{w h i t e}{.} {\text{g Fe"_3"O"_4 implies (y color(red)(cancel(color(black)("g"))))/(231.533color(red)(cancel(color(black)("g")))"mol"^(-1)) = (y/231.533)color(white)(.)"moles Fe"_3"O}}_{4}$

According to the balanced chemical equation, the reaction produces $1$ mole of iron(II, III) oxide for every $3$ moles of iron metal that it consumes.

This means that you have

"moles of Fe"/("moles of Fe"_3"O"_4) = 3

In your case, this will be equivalent to

$\left(\left(\frac{x}{55.845}\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/((y/231.533)color(red)(cancel(color(black)("moles}}}}\right) = 3$

which gets you

$\frac{x}{y} \cdot \frac{231.533}{55.845} = 3$

$\frac{x}{y} = 0.7236 \text{ " " } \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(2\right)}$

You now have two equations with two unknowns, so use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(2\right)}$ to say that

$x = 0.7236 \cdot y$

and plug this into equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(1\right)}$ to get

$648 = 600 - 0.7236 \cdot y + y$

Solve for $y$ to find

$0.2764 \cdot y = 648 - 600$

$y = \frac{48}{0.2764} = 173.66$

This means that you have

$x = 0.7236 \cdot 173.66$

$x = 125.66$

Since $x$ represents the mass of iron metal that reacted, you can say that the reaction consumed

$\text{mass of Fe that reacted" = "125.66 g}$

This means that the percent of iron metal that took place in the reaction is equal to

color(darkgreen)(ul(color(black)("% Fe that reacted"))) = (125.66 color(red)(cancel(color(black)("g"))))/(600color(red)(cancel(color(black)("g")))) xx 100% = color(darkgreen)(ul(color(black)(21%)))#

I'll leave the answer rounded to two sig figs, but keep in mind that you have only one significant figure for the mass of iron metal.