Question

Consider reacting copper (II) sulfate with iron. Two possible reactions can occur, as represented by the following equations.

$\begin{array}{l}\mathbf{Copper}\mathbf{\left(}\mathbf{II}\mathbf{\right)}\mathbf{sulfate}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{iron}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{copper}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{iron}\mathbf{\left(}\mathbf{II}\mathbf{\right)}\mathbf{sulfate}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\\ \mathbf{Copper}\mathbf{\left(}\mathbf{II}\mathbf{\right)}\mathbf{sulfate}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{iron}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{copper}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{iron}\mathbf{\left(}\mathbf{III}\mathbf{\right)}\mathbf{sulfate}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\end{array}$

You place $\mathbf{87}\mathbf{.}\mathbf{7}\mathbf{mL}$ of a $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{M}$ solution of copper$\mathbf{\left(}\mathbf{II}\mathbf{\right)}$ sulfate in a beaker. You then add $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{g}$ of iron filings to the copper $\mathbf{\left(}\mathbf{II}\mathbf{\right)}$ sulfate solution. After the reaction occurs, you isolate $\mathbf{2}\mathbf{.}\mathbf{27}\mathbf{g}$ of copper. Which equation above describes the reaction that occurred? Support your answer.

Short answer

The equation describes the reaction that takes place in reaction $1$is: $\mathrm{Copper}\left(\mathrm{II}\right)\mathrm{sulfate}\left(\mathrm{aq}\right)+\mathrm{iron}\left(\mathrm{s}\right)\to \mathrm{copper}\left(\mathrm{s}\right)+\mathrm{iron}\left(\mathrm{II}\right)\mathrm{sulfate}\left(\mathrm{aq}\right)$

Step-by-step solution

Introduction

Mole is a SI unit for measuring the amount of a material. The mole is the amount of a material that contains approximately the same number of atoms as $\mathbf{12}\mathbf{.}\mathbf{00}\mathbf{g}$ of $\mathbf{Carbon}\mathbf{-}\mathbf{12}$ .

The number of moles of a substance is defined as the ratio of the compound's given mass to its molar or molecular mass.

The mathematical formula is as follows:

$\mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{compound}}{\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{compound}}$

Explanation

The volume of copper$\left(\mathrm{II}\right)$ sulfate$=87.7\mathrm{mL}$

The concentration of coppersulfate$=0.500\mathrm{M}$

Mass of iron$=2.00\mathrm{g}$

Mass of copper$=2.27\mathrm{g}$

Calculating the number of moles

The calculation of themoles is:

$\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{copper}\left(\mathrm{II}\right)\mathrm{sulfate}=\mathrm{Molarity}\times \mathrm{Volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{L}$

Converting the volume into liters:.

$\begin{array}{c}\mathrm{Volume}\mathrm{in}\mathrm{L}=87.7\mathrm{mL}\times \frac{1\mathrm{L}}{1000\mathrm{mL}}\\ =0.0877\mathrm{L}\end{array}$

Thus,

$\begin{array}{c}\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{copper}\left(\mathrm{II}\right)\mathrm{sulfate}=0.500\mathrm{moles}/\mathrm{L}\times 0.0877\mathrm{L}\\ =0.0439\mathrm{mole}\end{array}$

Mass of copper isolated

Let iron be a limiting agent.

From reaction :

$\mathrm{Copper}\left(\mathrm{II}\right)\mathrm{sulfate}\left(\mathrm{aq}\right)+\mathrm{iron}\left(\mathrm{s}\right)\to \mathrm{copper}\left(\mathrm{s}\right)+\mathrm{iron}\left(\mathrm{II}\right)\mathrm{sulfate}\left(\mathrm{aq}\right)$

$\begin{array}{c}\mathrm{Mass}\mathrm{of}\mathrm{copper}\mathrm{isolated}=0.0358\mathrm{mole}\mathrm{iron}\times \frac{1\mathrm{mole}\mathrm{Cu}}{1\mathrm{mole}\mathrm{iron}}\times 63.55\mathrm{g}/\mathrm{mole}\mathrm{of}\mathrm{Cu}\\ =\mathbf{2}\mathbf{.}\mathbf{28}\mathbf{g}\end{array}$

From reaction :

$\mathrm{Copper}\left(\mathrm{II}\right)\mathrm{sulfate}\left(\mathrm{aq}\right)+\mathrm{iron}\left(\mathrm{s}\right)\to \mathrm{copper}\left(\mathrm{s}\right)+\mathrm{iron}\left(\mathrm{III}\right)\mathrm{sulfate}\left(\mathrm{aq}\right)$

role="math" localid="1657265913372" $\begin{array}{c}\mathrm{Mass}\mathrm{of}\mathrm{copper}\mathrm{isolated}=0.0439\mathrm{mole}{\mathrm{CuSO}}_4\times \frac{3\mathrm{mole}\mathrm{Cu}}{3\mathrm{mole}{\mathrm{CuSO}}_4}\times 63.55\mathrm{g}/\mathrm{mole}\mathrm{Cu}\\ =\mathbf{2}\mathbf{.}\mathbf{79}\mathbf{g}\end{array}$

Interpretation

Thus, the mass of copper isolated from reaction fits best with the supplied data, which is $\mathbf{2}\mathbf{.}\mathbf{27}\mathbf{g}$, based on the above two values of copper isolated mass.