# Given the following, how many millimoles (mmol) of the non-limiting reactant will remain unused at the end of the reaction?

## If iron is oxidized to $F {e}^{+} 2$ by a copper(II) sulfate solution, and 0.485 grams of iron and 14.5 mL of 0.438M copper(II) sulfate react to form as much product as possible

Aug 22, 2016

$\text{2.33 mmoles}$

#### Explanation:

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

${\text{Fe"_ ((s)) + "CuSO"_ (4(aq)) -> "FeSO"_ (4(aq)) + "Cu}}_{\left(s\right)}$

Notice that iron and copper(II) sulfate react in a $1 : 1$ mole ratio, which means that your limiting reagent will be the reactant that has the smallest number of moles available.

Use the molar mass of iron

M_("M Fe") = "55.845 g mol"^(-1)

to calculate how many moles of iron you have in your sample

0.485 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "0.008685 moles Fe"

Since the problem wants you to find the number of millimoles of the excess reactant, convert the number of moles of iron to millimoles

0.008685 color(red)(cancel(color(black)("moles"))) * (10^3"mmol")/(1color(red)(cancel(color(black)("moles")))) = "8.685 mmoles Fe"

Now, the problem provides you with the molarity and volume of the copper(II) sulfate solution, so use this info to find the number of millimoles of copper(II) sulfate added to the reaction

14.5 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * (0.438color(red)(cancel(color(black)("moles CuSO"_4))))/(1color(red)(cancel(color(black)("L")))) * (10^3"mmol")/(1color(red)(cancel(color(black)("mole")))) = "6.351 mmoles CuSO"_4

As you can see, you have fewer millimoles of copper(II) sulfate than of iron metal, which means that copper(II) sulfate will be your limiting reagent.

On other words, iron metal will be in excess.

Since the reaction consumes the two reactants in a $1 : 1$ mole ratio, you can say that after the reaction is complete, your solution will contain

n_("CuSO"_4) = 6.351 - 6.351 = "0 mmoles CuSO"_4 -> completely consumed

n_("Fe") = 8.685 - 6.351 = "2.334 mmoles Fe"

Rounded to three sig figs, the answer will be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{no. of mmoles of Fe left} = 2.33} \textcolor{w h i t e}{\frac{a}{a}} |}}}$