Stoichiometry Notes

Proportional Relationships

  • Analogy: Cheese Sandwich
    • Recipe: 2 Bread (Bd) + 2 Cheese (Ch) + 1 Lettuce (L) → 1 Sandwich (Bd2Ch2L)
    • Question: If you have 4 Bd, 4 Ch, and 2 L, how many sandwiches can you make?

Stoichiometry

  • Definition: The ratios between coefficients in a chemical reaction, based on the mole ratio.
  • Example: 2Mg+O2→2MgO2 Mg + O_2 \rightarrow 2 MgO
  • Stoichiometric coefficients are used to determine mole ratios.

Stoichiometry Problems

  • Problem 1: How many moles of KClO3KClO_3 must decompose to produce 9 moles of oxygen gas?
  • Reaction: 2KClO<em>3→2KCl+3O</em>22KClO<em>3 \rightarrow 2KCl + 3O</em>2
  • Given: 9 moles of O2O_2
  • Calculation:
    9 mol O<em>2×2 mol KClO</em>33 mol O<em>2=6 mol KClO</em>39 \, mol \, O<em>2 \times \frac{2 \, mol \, KClO</em>3}{3 \, mol \, O<em>2} = 6 \, mol \, KClO</em>3
  • Start with the given number, use a ratio to cancel out unwanted units.

Stoichiometry Problems Continued

  • Reaction: CaCl<em>2→Ca+Cl</em>2CaCl<em>2 \rightarrow Ca + Cl</em>2
    • Problem: How many moles of calcium metal are produced from the decomposition of 8 moles of calcium chloride?
    • Calculation:
      8 mol CaCl<em>2×1 mol Ca1 mol CaCl</em>2=8 mol Ca8 \, mol \, CaCl<em>2 \times \frac{1 \, mol \, Ca}{1 \, mol \, CaCl</em>2} = 8 \, mol \, Ca
    • Problem: How many moles of calcium metal and chlorine gas are produced from the decomposition of 8 moles of calcium chloride?
    • Calculation:
      8 mol CaCl<em>2×1 mol Cl</em>21 mol CaCl<em>2=8 mol Cl</em>28 \, mol \, CaCl<em>2 \times \frac{1 \, mol \, Cl</em>2}{1 \, mol \, CaCl<em>2} = 8 \, mol \, Cl</em>2
    • Result: 8 moles of calcium metal and 8 moles of chlorine gas are produced.

More Stoichiometry Problems

  • Reaction: 2Sb+3Cl<em>2→2SbCl</em>32 Sb + 3 Cl<em>2 \rightarrow 2 SbCl</em>3
    • Problem: How many moles of chlorine gas are required to react with 5 moles of antimony?
    • Calculation:
      5 mol Sb×3 mol Cl<em>22 mol Sb=7.5 mol Cl</em>25 \, mol \, Sb \times \frac{3 \, mol \, Cl<em>2}{2 \, mol \, Sb} = 7.5 \, mol \, Cl</em>2
    • Problem: How many moles of SbCl<em>3SbCl<em>3 are produced from 5 moles of antimony and excess Cl</em>2Cl</em>2?
    • Calculation:
      5 mol Sb×2 mol SbCl<em>32 mol Sb=5 mol SbCl</em>35 \, mol \, Sb \times \frac{2 \, mol \, SbCl<em>3}{2 \, mol \, Sb} = 5 \, mol \, SbCl</em>3
    • Problem: How many moles of SbCl<em>3SbCl<em>3 are produced from 7.5 moles of Cl</em>2Cl</em>2 and excess Sb?
    • Calculation:
      7.5 mol Cl<em>2×2 mol SbCl</em>33 mol Cl<em>2=5 mol SbCl</em>37.5 \, mol \, Cl<em>2 \times \frac{2 \, mol \, SbCl</em>3}{3 \, mol \, Cl<em>2} = 5 \, mol \, SbCl</em>3

Stoichiometry with Larger Coefficients

  • Reaction: 2C<em>6H</em>6+15O<em>2→12CO</em>2+6H2O2C<em>6H</em>6 + 15O<em>2 \rightarrow 12 CO</em>2 + 6H_2O
    • Problem: How many moles of oxygen gas are required to react with 5.00 moles of C<em>6H</em>6C<em>6H</em>6?
    • Calculation:
      5 mol C<em>6H</em>6×15 mol O<em>22 mol C</em>6H<em>6=37.5 mol O</em>25 \, mol \, C<em>6H</em>6 \times \frac{15 \, mol \, O<em>2}{2 \, mol \, C</em>6H<em>6} = 37.5 \, mol \, O</em>2
    • Problem: How many moles of CO<em>2CO<em>2 are produced from 5 moles of O</em>2O</em>2 and excess C<em>6H</em>6C<em>6H</em>6?
    • Calculation:
      5 mol O<em>2×12 mol CO</em>215 mol O<em>2=4 mol CO</em>25 \, mol \, O<em>2 \times \frac{12 \, mol \, CO</em>2}{15 \, mol \, O<em>2} = 4 \, mol \, CO</em>2