Consider reacting copper(II) sulfate with iron. Two possible reactions can occur, as represented by the following equations. \(\operatorname{copper}(\) II \()\) sulfate \((a q)+\) iron \((s) \longrightarrow\) \(\operatorname{copper}(s)+\) iron(II) sulfate \((a q)\) copper(II) sulfate \((a q)+\operatorname{iron}(s) \longrightarrow\) \(\operatorname{copper}(s)+\) iron(III) sulfate \((a q)\) You place \(87.7 \mathrm{~mL}\) of a \(0.500-M\) solution of copper(II) sulfate in a beaker. You then add \(2.00 \mathrm{~g}\) of iron filings to the copper(II) sulfate solution. After one of the above reactions occurs, you isolate \(2.27 \mathrm{~g}\) of copper. Which equation above describes the reaction that occurred? Support your answer.

Step by step solution

01

Identify the given information

In the problem, we are given: - 87.7 mL of 0.500-M copper(II) sulfate (CuSO4) solution - 2.00 g of iron (Fe) filings - 2.27 g of the isolated copper (Cu) We are also given two possible reactions: 1. \(CuSO_4(aq)+Fe(s) → Cu(s)+FeSO_4(aq)\) 2. \(CuSO_4(aq)+Fe(s) → Cu(s)+2Fe_2(SO_4)_3(aq)\)

02

Calculate the moles of copper(II) sulfate and iron

To find the moles of the reactants, we can use the molarity formula and the mass-to-mole conversion: 1. Calculate the moles of copper(II) sulfate (CuSO4): Moles of CuSO4 = Molarity × Volume in liters Moles of CuSO4 = 0.500 M × 0.0877 L = 0.04385 mol 2. Calculate the moles of iron (Fe): Moles of Fe = mass / molar mass of Fe Moles of Fe = 2.00 g / 55.845 g/mol = 0.03580 mol

03

Determine the limiting reactant for both possible reactions

Now we can find the limiting reactant for each reaction using mole ratios from the balanced chemical equations: For Reaction 1: Ratio CuSO4/Fe = 1 Moles of CuSO4 = 0.04385 mol Moles of Fe = 0.03580 mol In this case, Fe will be limiting the reactant since we have less moles compared to CuSO4. For Reaction 2: Notice that the balanced equation for reaction 2 should be: \(CuSO_4(aq) + 2Fe(s) \rightarrow Cu(s) + Fe_2(SO_4)_3(aq)\) Ratio CuSO4/Fe = 1/2 Moles of CuSO4 = 0.04385 mol Moles of Fe = 0.03580 mol In this case, CuSO4 will be the limiting reactant since we need two times more moles of Fe for the reaction to occur.

04

Calculate the mass of copper produced in both reactions

We will now calculate the mass of copper produced, given the limiting reactant: For Reaction 1: The mole ratio of CuSO4 to Cu is 1:1. Therefore, molar mass of Cu = molar mass of CuSO4 = 0.03580 mol Mass of Cu = moles of Cu × molar mass of Cu Mass of Cu = 0.03580 mol × 63.546 g/mol = 2.27 g For Reaction 2: The mole ratio of CuSO4 to Cu is 1:1. Therefore, molar mass of Cu = molar mass of CuSO4 = 0.04385 mol Mass of Cu = moles of Cu × molar mass of Cu Mass of Cu = 0.04385 mol × 63.546 g/mol = 2.79 g

05

Conclude which reaction occurred

We are given the experimental mass of copper to be 2.27 g. Comparing this value with the theoretical values obtained from both reactions, we can see that the mass of copper from Reaction 1 (2.27 g) matches the experimental value. Therefore, the first reaction: \(CuSO_4(aq) + Fe(s) \rightarrow Cu(s) + FeSO_4(aq)\) is the reaction that occurred, as it supports the given information.

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