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8.3: Stoichiometry of Aqueous Reactions

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    283932
  • Introduction

    At the beginning of this Chapter we covered stoichiometry of reactions where everything was a solid, and we used the mass and formula weight to calculate the moles of reactants or products consumed or produced. For a reaction where two chemicals interact with each other they must come into contact with each other if the reaction is going to proceed. One way of enabling this mobility is to dissolve the reactants into a solvent creating a solution. Unless you have a pure substance, you can not use mass to measure the number of particles, as you are weighting both the solvent and solute(s). So if you have a solution, you need to know the concentration, and the most common way of representing chemical concentrations is the molarity.

    Solid phase:
    \[ \begin{align}n(moles) & = \frac{m(g)}{fw(\frac{g}{mol} }\\ \textcolor{red}{n} & \textcolor{red}{=\frac{m}{fw}} \end{align}\]

    Aqueous phase (solute concentration):
    \[\begin{align} n(moles) & = M\left ( \frac{mol}{L} \right )V(L) \\ \textcolor{red}{n} & \textcolor{red}{=MV} \end{align}\]

    where,
    n = moles
    m = mass
    V = Volume
    fw = formula weight (molar mass)

    So now we can calculate the moles of a pure substance if we weigh it's mass and know it's molar mass, or if we measure its volume and know it's concentration (Molarity).

    If given a solid, you need to know its molar mass

    clipboard_e5dc9173f93d027f9b235a33445f2547f.png

    If given a solute, you need to know its concentration (molarity)

    clipboard_e85612b2e05a7fe0195a04ef3fbca0fab.png

    Solution Stoichiometry

    In sections 4.1 and 4.2 we calculated theoretical yield, limiting reagent and excess reagents for reactions where all chemical species were described as solids where we measured the mass. Now that we have a way to describe solute concentrations (Molarity) we can extend these to include solutions where we measure the volume. Lets look at the reaction of adding a solid soluble salt to an aqueous solution

    Dissolving 10.0 g of Copper(II)sulfate into 500.0 mL of 0.100 M sodium phosphate

    Exercise \(\PageIndex{1}\)

    What are the formulas of the reactants when 10 g of copper(II) sulfate is dissolved into 500.0 mL of 0.100 M sodium phosphate? (see video \(\PageIndex{1}\))

    Answer

    CuSO4(s) and Na3PO4(aq)

    Exercise \(\PageIndex{2}\)

    What are the formulas and phases of the products when 10 g of copper(II) sulfate is dissolved into 500.0 mL of 0.100 M sodium phosphate?(see video \(\PageIndex{1}\))

    Answer

    Na2SO4(aq) and Cu3(PO4)2(s)

    Exercise \(\PageIndex{3}\)

    Write the balanced equation when 10 g of copper(II) sulfate is dissolved into 500.0 mL of 0.100 M sodium phosphate?(see video \(\PageIndex{1}\))

    Answer

    3CuSO4(aq) + 2Na3PO4(aq) --> 3Na2SO4(aq) and Cu3(PO4)2(s)

    NOTE: We added copper(II)sulfate as a solid, but it dissolved (see solubility rules, section 3.4.2.1), and so it should be treated as aqueous when writing the equation

    Video: \(\PageIndex{1}\): 3'54" YouTube identifying reactants, products, phases and balancing the equation for the dissolution of 10 g of copper(II) sulfate into 500.0 mL of 0.100 M sodium phosphate (https://youtu.be/qdtiFATDMdU).

    Calculating the Theoretical Yield for the mass of precipitate formed

    Example \(\PageIndex{4}\)

    Calculate the theoretical yield for the mass of precipitate formed when 10.0 g of copper(II) sulfate is dissolved into 500.0 mL of 0.100 M sodium phosphate.

    Solution

    See video \(\PageIndex{2}\) for the solution of this problem, and video \(\PageIndex{3}\) for a deeper explanation of the fast technique used in video 2. Note how we set up the equation by writing out the given data under each species, along with the moles that we need to calculate with that data. We calculate for the moles of each reactant (n), divide by its storichiometric coefficient, and the smallest number is the limiting reagent. We then proceed and calculate the mass of copper(II)phosphate formed.

    \[\underset{n= \\ m=10.0g \\ fw=159.61g/mol}{3CuSO_4(s)} + \underset{n= \\ M=0.100mol/l \\ V=500.0ml}{2Na_3PO_4(aq)} \rightarrow \underset{m=? \\ fw=380.58g/mol}{Cu_3(PO_4)_2(s)} + 3Na_2SO_4(aq) \]

    at 2'25" into video \(\PageIndex{2}\) we calculated the limiting reagent, which was copper(II)Sulfate and thus based the yield on its consumption. The overall equation is

    \[10.0gCuSO_4(s)\left ( \frac{molCuSO_4}{159.69g} \right )\left ( \frac{1molCu_3(PO_4)_2}{3molCuSO_4} \right )\left ( \frac{380.58gCu_3(PO_4)_2(s)}{mol} \right )=7.95gCu_3(PO_4)_2(s)\]

    Note, the above equation is not used in the fast technique, but is used in video \(\PageIndex{3}\) when we explain the fast technique

    Video: \(\PageIndex{2}\): 3'30" YouTube video calculating the mass precipitate formed by complete consumption of the limiting reagent for the dissolution of 10 g of copper(II) sulfate into 500.0 mL of 0.100 M sodium phosphate (https://youtu.be/Vh3Je39bgQc)
    Video: \(\PageIndex{3}\): The fast technique in video \(\PageIndex{2}\) explained in more detail (https://youtu.be/5Qlo1bi1uzE).

    Calculating excess reagents

    Since the copper sulfate is the limiting reagent we can calculate the amount of excess sodium phosphate

    Example \(\PageIndex{2}\): Excess Reagent

    Calculate the concentration (molarity) of excess phosphate when 10 g of copper(II) sulfate is dissolved into 500.0 mL of 0.100 M sodium phosphate (note: the real reaction is expressed by the net ionic equation:

    \[3Cu^{+2}(aq) + 2PO_4^{-3}(aq) \rightarrow Cu_3(PO_4)_2(s)\]

    Solution

    see Video \(\PageIndex{4}\)

    First calculate moles excess phosphate

    \[\text{Initial moles} - \text{amount consumed by reaction with limiting reagent} \\ 0.050 mol PO_4^{-3} -0.06265mol Cu^{+2}\left ( \frac{2 \; mol \; PO_4^{-3}}{3 \; mol \; Cu^{+2}} \right )=0.0082mol PO_4^{-3}\]

    then divide into total volume

    \(\frac{0.0082mol PO_4^{-3}}{0.500l}= 0.016MPO_4^{-3}(aq)\)

    Video \(\PageIndex{4}\): Determining the excess phosphate when 10 g of copper(II) sulfate is dissolved into 500.0 mL of 0.100 M sodium phosphate https://youtu.be/xsp1_BADnqA).

    Calculating Concentration of Spectator Ions

    For the reaction that occurs when 10 g of copper(II) sulfate is dissolved into 500.0 mL of 0.100 M sodium phosphate we have calculated the theoretical yield

    Example \(\PageIndex{3}\) Spectator Ion Concentrations

    Calculate spectator ion concentrations when 10 g of copper(II) sulfate is dissolved into 500.0 mL of 0.100 M sodium phosphate

    Solution

    See video \(\PageIndex{5}\) for the solution to this problem. There are several ways you can set this up. In calculating the limiting reagent we determined the moles of each reactant and so that would be a logical place to start

    \[\underset{n=\frac{m(g)}{fw(g/mol)}=\frac{10.0g}{159.61g/m0l} \\ n=0.06265mol}{3CuSO_4(s)} + \underset{n=MV=(0.100mol/l)(0.5000l \\ n = 0.0500mol}{Na_3PO_4(aq)} \rightarrow Cu_3(PO_4))2(s) + 3Na_2SO_4(aq) \]

    \[\frac{molSO_4^{-2}}{V_{total}}=\frac{0.06265molCuSO_4\left ( \frac{1 molSO_4^{-2}}{molCuSO_4} \right )}{0.500l}=0.125M\]

    \[\frac{molNa^+}{V_{total}}=\frac{0.0500molNa_3PO_4\left ( \frac{3 molNa^+}{molNa_3PO_4} \right )}{0.500l}=0.300M\]

    Video \(\PageIndex{5}\): 2'44" YouTube calculating the spectator ion concentration when 10 g of copper(II) sulfate is dissolved into 500.0 mL of 0.100 M sodium phosphate.

    Test Yourself

    Homework: Section 8.3

    Graded Assignment: Section 8.3

    Exercise \(\PageIndex{4}\)

    Consider the reaction of sodium phosphate and barium chloride, where 25.00 ml of 0.8000M Na3PO4 is mixed with 25.00ml of 0.1000M BaCl2 (and the final volume is 50.00 ml, that is there is no change in volume upon mixing).

    1. What is the limiting reagent?
    2. What is the identity and mass of precipitate formed?
    3. What is the concentration of the sodium spectator ion?
    4. What is the concentration of the excess phosphate?
    Answer a

    \(BaCl_2\)

    Answer b
    \(0.5016g Ba_3(PO_4)_2 \\ 0.002500molBaCl_2\left ( \frac{1molBa_3(PO_4)_3)}{3molBaCl_2} \right )\left ( \frac{601.93g Ba_3(PO_4)_2)}{mol} \right )=0.5016g Ba_3(PO_4)_2\)
    Answer c

    0.1200M

    \(M_fV_f=M_iV_i \\
    M_F=\frac{M_iV_i}{V_f} \\
    \frac{0.8000mol/L \; Na_3PO_4\left ( \frac{3molNa^+}{molNa_3PO_4} \right )0.02500L \; Na_3PO_4}{0.02500L + 0.02500L}=1.200MNa^+\)

    Answer d

    \[mol_{excess}PO_4^{-3} = mole_{initial}-mol_{consumed} \nonumber \\ 0.8000molNa_3PO_4/L\left ( \frac{1molPO_4^{-3}}{molNa_3PO_4} \right )0.02500L-\left ( 0.100mol/L \; BaCl_2\left ( \frac{1molBa^{+2}}{molBaCl_2} \right )0.02500L\left ( \frac{2molPO_4^{-3}}{3molBa^{+2}} \right ) \right )=0.01833molPO_4^{-3}\\ Concentration=\frac{mol_{excess}PO_4^{-3}}{V_T}=\frac{0.01833molPO_4^{-3}}{0.02500L + 0.02500L}=0.3667M \; PO_4^{-3}\]

    Contributors and Attributions

    Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:

    • Ronia Kattoum (UALR)
    • anonymous
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