# 385mg of iron reacts with excess bromine, producing 1921mg of a mixture of FeBr2 and FeBr3. How would you calculate the masses of FeBr2 and FeBr3 in the product mixture?

Nov 21, 2015

Here's what I got.

#### Explanation:

You know that $\text{385 mg}$ of iron will react with excess bromine to produce a mixture of iron(II) bromide, ${\text{FeBr}}_{2}$, and iron(III) bromide, ${\text{FeBr}}_{3}$.

The total mass of the mixture is equal to $\text{1921 mg}$.

Your strategy here will be to write the balanced chemical equations for these two reactions, then use the mole ratios that exist between iron and the two products to determine how many moles of each were produced by the reaction.

${\text{Fe"_text((s]) + "Br"_text(2(l]) -> "FeBr}}_{\textrm{2 \left(s\right]}}$

and

$2 {\text{Fe"_text((s]) + 3"Br"_text(2(l]) -> 2"FeBr}}_{\textrm{3 \left(s\right]}}$

Notice that in both cases, you have a $1 : 1$ mole ratio between iron and the product of the reaction.

Now, the total number of moles of iron available for the reactions will be

835color(red)(cancel(color(black)("mg"))) * (1color(red)(cancel(color(black)("g"))))/(1000color(red)(cancel(color(black)("mg")))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "0.006894 moles Fe"

Now, let's assume that the number of moles of iron that react to form iron(II) bromide is equal to $x$, and the number of moles of iron that react to form iron(III) bromide is equal to $y$.

You know that you have the following relationship between $x$ and $y$

$x + y = 0.006894 \text{ " " } \textcolor{p u r p \le}{\left(1\right)}$

You can use the aforementioned mole ratios to say that the first reaction will produce $x$ moles of iron(II) bromide and that the second reaction will produce $y$ moles of iron(III) bromide.

Next, use the respective molar masses of the two products to find a relationship between how many moles and how many grams were produced by the reaction.

• iron(II) bromide $\to \text{215.65 g/mol}$
• iron(III) bromide $\to \text{295.56 g/mol}$

The mass of iron(II) bromide produced by the first reaction will be

xcolor(red)(cancel(color(black)("moles"))) * "215.65 g"/(1color(red)(cancel(color(black)("mole")))) = x * "215.65 g"

The mass of iron(III) bromide produced by the second rection will be

ycolor(red)(cancel(color(black)("moles"))) * "295.56 g"/(1color(red)(cancel(color(black)("mole")))) = y * "295.56 g"

Since you know that the total mass of the mixture is equal to $\text{1921 mg}$, you can say that - remember to convert the total mass to grams!

$x \cdot 216.65 + y \cdot 295.56 = 1.921 \text{ " " } \textcolor{p u r p \le}{\left(2\right)}$

Use equation $\textcolor{p u r p \le}{\left(1\right)}$ to get

$x = 0.006894 - y$

Plug this into equation $\textcolor{p u r p \le}{\left(2\right)}$ to get

$215.65 \cdot \left(0.006894 - y\right) + y \cdot 295.56 = 1.921$

$1.4867 - 215.65 \cdot y + 295.56 \cdot y = 1.921$

$79.91 \cdot y = 0.4343 \implies y = 0.005435$

Plug this value of $y$ into equation $\textcolor{p u r p \le}{\left(1\right)}$ to find the value of $x$

$x = 0.006894 - 0.005431 = 0.001459$

So, the two reactions will produce

• $\text{0.001459 moles}$ of iron(II) bromide
• $\text{0.005435 moles}$ of iron(III) bromide

The masses of the two products will thus be

${m}_{F e B {r}_{2}} = 0.001459 \cdot {\text{215.65 g" = "0.315 g FeBr}}_{2}$

${m}_{F e B {r}_{3}} = 0.005435 \cdot {\text{295.56 g" = "1.606 g FeBr}}_{3}$

Expressed in milligrams, the answer will be

${m}_{F e B {r}_{2}} = \textcolor{g r e e n}{\text{315 mg}}$

${m}_{F e B {r}_{3}} = \textcolor{g r e e n}{\text{1606 mg}}$