Solution Stoichiometry

Related Examples and Practice Problems

Additional Worked Out Examples/ Practice

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  • Relating Properties to Composition: Predicting classification based on descriptive properties

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Core Concept

Solution stoichiometry is the study of quantitative relationships in chemical reactions that occur in solution. It combines the principles of stoichiometry with solution concentrations to calculate the amounts of reactants and products in reactions involving dissolved substances.

Practice Tips

  • Always Balance the Equation First: Correct mole ratios are essential for solution stoichiometry.

  • Convert Volumes to Liters: Ensure all volumes are in liters when using molarity.

  • Use Molarity for Mole Conversions: Molarity is the key to finding moles from solution volumes.

  • Be Mindful of Units: Watch for conversions between liters, milliliters, grams, and moles.

  • Practice with Different Reaction Types: Solution stoichiometry applies to neutralization, precipitation, and redox reactions.

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Key Concepts

Molarity (M):

  • Molarity is the concentration of a solution, defined as the moles of solute per liter of solution:

  • $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

  • Units are in moles per liter (mol/L \text{mol/L}mol/L), also written as MMM.

Mole Ratios:

  • As in standard stoichiometry, the coefficients in a balanced equation provide mole ratios, which relate the moles of reactants to the moles of products.

  • These mole ratios are essential for converting between moles of different substances in a reaction.

Volume Relationships:

  • Solution stoichiometry problems often involve calculating volumes of solutions required to react with a given amount of a substance, or the volume of solution needed to achieve a desired concentration.

Dilution:

  • In many stoichiometry problems, solutions are diluted to a lower concentration.

  • The dilution formula is: $M_1 V_1 = M_2 V_2$

  • Where $M_1$​ and $V_1$​ are the initial molarity and volume, and $M_2$​ and $V_2$​ are the final molarity and volume.

Steps for Solution Stoichiometry Problems

  1. Write and Balance the Chemical Equation:

    • Ensure that the reaction is balanced so that you can use accurate mole ratios.

  2. Convert Volumes and Concentrations to Moles:

    • Use molarity to convert between moles and volume of a solution: $\text{moles} = M \times V$

    • Ensure volume is in liters; if given in milliliters, convert to liters by dividing by 1000.

  3. Use Mole Ratios to Find Unknown Quantities:

    • Use the balanced equation’s mole ratios to relate the moles of the known substance to the moles of the unknown substance.

  4. Convert Moles Back to Desired Units:

    • If needed, convert moles back to volume using the molarity, or to grams using molar mass.

Example Problem: Solution Stoichiometry in a Neutralization Reaction

Problem: How many milliliters of 0.500 M HCl are needed to react completely with 25.0 mL of 0.250 M NaOH?

HCl + NaOH → NaCl + H₂O

Solution:

  1. Write and Balance the Equation:

    • The equation is already balanced as written.

  2. Calculate Moles of NaOH:

    • Molarity of NaOH = 0.250 M, volume = 25.0 mL (0.0250 L)

    • Moles of NaOH = 0.250 M * 0.0250 L = 0.00625 mol

  3. Use Mole Ratio to Find Moles of HCl:

    • The balanced equation shows a 1:1 mole ratio between HCl and NaOH.

    • Moles of HCl = 0.00625 mol

  4. Convert Moles of HCl to Volume:

    • Molarity of HCl = 0.500 M

    • Volume of HCl = (Moles of HCl) / (Molarity of HCl) = (0.00625 mol) / (0.500 M) = 0.0125 L = 12.5 mL

Answer: 12.5 mL of 0.500 M HCl is needed.

Example Problem: Precipitation Reaction in Solution Stoichiometry

Problem: What mass of AgCl precipitate will form when 50.0 mL of 0.100 M AgNO₃ is mixed with excess NaCl solution?

AgNO₃ + NaCl → AgCl + NaNO₃

Solution:

  1. Write and Balance the Equation:

    • The equation is balanced as written.

  2. Calculate Moles of AgNO₃:

    • Molarity of AgNO₃ = 0.100 M, volume = 50.0 mL (0.0500 L)

    • Moles of AgNO₃ = 0.100 M * 0.0500 L = 0.00500 mol

  3. Use Mole Ratio to Find Moles of AgCl:

    • The balanced equation shows a 1:1 mole ratio between AgNO₃ and AgCl.

    • Moles of AgCl = 0.00500 mol

  4. Convert Moles of AgCl to Grams:

    • Molar mass of AgCl = 143.32 g/mol

    • Mass of AgCl = 0.00500 mol * 143.32 g/mol = 0.7166 g

Answer: 0.7166 g of AgCl will form.

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