# 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?

##### 1 Answer

Here's what I got.

#### Explanation:

You know that

The total mass of the mixture is equal to

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.

#"Fe"_text((s]) + "Br"_text(2(l]) -> "FeBr"_text(2(s])#

and

#2"Fe"_text((s]) + 3"Br"_text(2(l]) -> 2"FeBr"_text(3(s])#

Notice that in both cases, you have a

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

You know that you have the following relationship between

#x + y = 0.006894" " " "color(purple)((1))#

You can use the aforementioned mole ratios to say that the first reaction will produce

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#-> "215.65 g/mol"# iron(III) bromide#-> "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 *grams*!

#x * 216.65 + y * 295.56 = 1.921" " " "color(purple)((2))#

Use equation

#x = 0.006894 - y#

Plug this into equation

#215.65 * (0.006894 - y) + y * 295.56 = 1.921#

#1.4867 - 215.65 * y + 295.56 * y = 1.921#

#79.91 * y = 0.4343 implies y = 0.005435#

Plug this value of

#x = 0.006894 - 0.005431 = 0.001459#

So, the two reactions will produce

#"0.001459 moles"# of iron(II) bromide#"0.005435 moles"# of iron(III) bromide

The **masses** of the two products will thus be

#m_(FeBr_2) = 0.001459 * "215.65 g" = "0.315 g FeBr"_2#

#m_(FeBr_3) = 0.005435 * "295.56 g" = "1.606 g FeBr"_3#

Expressed in *milligrams*, the answer will be

#m_(FeBr_2) = color(green)("315 mg")#

#m_(FeBr_3) = color(green)("1606 mg")#