Stoichiometry and Chemical Reaction Calculations

Introduction to Stoichiometry

  • Stoichiometry Definition: Stoichiometry is defined as the calculation of quantities of any substances involved in a chemical reaction from the known quantities of other substances present in that reaction.

  • Foundational Requirements: Stoichiometry refers specifically to the ratios of substances in a chemical reaction. Because it relies on these ratios, a balanced chemical equation is strictly required to perform calculations.

  • Interpreting Chemical Equations: A balanced equation, such as the reaction between phosphorus and chlorine gas, provides Information on two levels:

    • Microscopic (Molecular) Level: 22 atoms of phosphorus react with 33 molecules of Cl2Cl_2 to produce 22 molecules of PCl3PCl_3.

    • Macroscopic (Molar) Level: 22 moles of phosphorus react with 33 moles of Cl2Cl_2 to produce 22 moles of PCl3PCl_3.

  • Chemical Reaction Example: 2P(s)+3Cl2(g)→2PCl3(l)2 P(s) + 3 Cl_2(g) \rightarrow 2 PCl_3(l).

Mole Calculations for Chemical Reactions (Section 5.1)

  • Core Goal: To calculate the number of moles of any substance involved in a chemical reaction based on the number of moles of any other substance provided in the reaction.

  • Flowchart of Mole Conversions:

    • Formula Units of A can be converted to Moles of A using Avogadro's Number (6.022×10236.022 \times 10^{23}).

    • Moles of A can be converted to Moles of B using the ratios obtained from the Balanced Chemical Equation.

    • Moles of B can be converted to Formula Units of B using Avogadro's Number.

  • Reacting Ratios: Using the equation 2P(s)+3Cl2(g)→2PCl3(l)2 P(s) + 3 Cl_2(g) \rightarrow 2 PCl_3(l), the following six possible mole ratios can be derived:

    • 3 mol Cl22 mol P\frac{3\,mol\,Cl_2}{2\,mol\,P}

    • 3 mol Cl22 mol PCl3\frac{3\,mol\,Cl_2}{2\,mol\,PCl_3}

    • 2 mol P2 mol PCl3\frac{2\,mol\,P}{2\,mol\,PCl_3}

    • 2 mol P3 mol Cl2\frac{2\,mol\,P}{3\,mol\,Cl_2}

    • 2 mol PCl33 mol Cl2\frac{2\,mol\,PCl_3}{3\,mol\,Cl_2}

    • 2 mol PCl32 mol P\frac{2\,mol\,PCl_3}{2\,mol\,P}

  • Example 5.2: Calculating Moles from Moles:

    • Problem: Calculate the number of moles of aluminum (AlAl) that will react with 3.18 mol3.18\,mol of oxygen (O2O_2) to form aluminum oxide (Al2O3Al_2O_3).

    • Balanced Equation: 4Al(s)+3O2(g)→2Al2O3(s)4 Al(s) + 3 O_2(g) \rightarrow 2 Al_2O_3(s)

    • Calculation:         3.18 mol O2×(4 mol Al3 mol O2)=4.24 mol Al3.18\,mol\,O_2 \times \left( \frac{4\,mol\,Al}{3\,mol\,O_2} \right) = 4.24\,mol\,Al

  • Example 5.4: Calculations with Partial Decomposition:

    • Problem: A sample of 0.1712 mol0.1712\,mol solid KClO3KClO_3 is heated. During the process, 0.1146 mol0.1146\,mol of the compound decomposes into KClKCl and O2O_2 gas. Calculate the number of moles of oxygen gas produced.

    • Step 1: Write Balanced Equation: 2KClO3(s)→3O2(g)+2KCl(s)2 KClO_3(s) \rightarrow 3 O_2(g) + 2 KCl(s)

    • Step 2: Identify Reacting Quantity: Although 0.1712 mol0.1712\,mol is present, only the amount that actually decomposes (0.1146 mol0.1146\,mol) is used for the stoichiometric calculation.

    • Calculation:         0.1146 mol KClO3×(3 mol O22 mol KClO3)=0.1719 mol O20.1146\,mol\,KClO_3 \times \left( \frac{3\,mol\,O_2}{2\,mol\,KClO_3} \right) = 0.1719\,mol\,O_2

Mass Calculations for Chemical Reactions (Section 5.2)

  • Foundational Principle: Reacting ratios provided by chemical equations are mole ratios, not mass ratios.

  • Molar Mass Utility: Molar mass must be used to convert masses of substances into moles before the stoichiometric mole ratio can be applied, and then used again to convert the resulting moles back into grams if required.

  • Flowchart of Mass and Mole Conversions:

    • Mass of Substance A →\rightarrow Moles of Substance A (using Molar Mass of A).

    • Moles of Substance A →\rightarrow Moles of Substance B (using Balanced Chemical Equation Coefficients).

    • Moles of Substance B →\rightarrow Mass of Substance B (using Molar Mass of B).

  • Example 5.6: Mass-to-Mass Calculation (Electrolysis):

    • Problem: Calculate the mass of chlorine gas (Cl2Cl_2) in grams produced by the electrolysis of 50.0 kg50.0\,kg of sodium chloride (NaClNaCl) in concentrated aqueous solution.

    • Balanced Equation: 2NaCl(aq)+2H2O(l)→electricity2NaOH(aq)+Cl2(g)+H2(g)2 NaCl(aq) + 2 H_2O(l) \xrightarrow{\text{electricity}} 2 NaOH(aq) + Cl_2(g) + H_2(g)

    • Unit Conversion: 50.0 kg NaCl=50,000 g NaCl50.0\,kg\,NaCl = 50,000\,g\,NaCl

    • Molar Masses: NaCl=58.44 g/molNaCl = 58.44\,g/mol; Cl2=70.903 g/molCl_2 = 70.903\,g/mol.

    • Dimensional Analysis:         50.0 kg NaCl×(1000 g NaCl1 kg NaCl)×(1 mol NaCl58.44 g NaCl)×(1 mol Cl22 mol NaCl)×(70.903 g Cl21 mol Cl2)=3.03×104 g Cl250.0\,kg\,NaCl \times \left( \frac{1000\,g\,NaCl}{1\,kg\,NaCl} \right) \times \left( \frac{1\,mol\,NaCl}{58.44\,g\,NaCl} \right) \times \left( \frac{1\,mol\,Cl_2}{2\,mol\,NaCl} \right) \times \left( \frac{70.903\,g\,Cl_2}{1\,mol\,Cl_2} \right) = 3.03 \times 10^4\,g\,Cl_2

  • Example 5.7: Industrial Scale Calculations:

    • Problem: Calculate the number of moles of SO2SO_2 gas required to prepare 50.050.0 metric tons of liquid H2SO4H_2SO_4.

    • Constants: 1 metric ton=1×106 g1\,metric\,ton = 1 \times 10^6\,g.

    • Balanced Equation: 2SO2(g)+O2(g)+2H2O(l)→2H2SO4(l)2 SO_2(g) + O_2(g) + 2 H_2O(l) \rightarrow 2 H_2SO_4(l)

    • Molar Mass of H2SO4H_2SO_4: 98.08 g/mol98.08\,g/mol.

    • Calculation:         50.0 tons H2SO4×(1×106 g1 ton)×(1 mol H2SO498.08 g H2SO4)×(2 mol SO22 mol H2SO4)=5.10×105 mol SO250.0\,tons\,H_2SO_4 \times \left( \frac{1 \times 10^6\,g}{1\,ton} \right) \times \left( \frac{1\,mol\,H_2SO_4}{98.08\,g\,H_2SO_4} \right) \times \left( \frac{2\,mol\,SO_2}{2\,mol\,H_2SO_4} \right) = 5.10 \times 10^5\,mol\,SO_2