chem 2/14

Balancing Chemical Equations

• To balance a chemical equation, count the number of each type of atom on both sides and adjust coefficients to make them equal.

• For the example:

  • C6H12O6 (glucose) decomposing with O2:

    • Six carbons require 6 CO2.

    • 12 hydrogens from 6 H2O require 6 O2.

    • Total O2 needed = 12 from CO2 + 6 from H2O = 18.

    • Balance O2 by placing a coefficient of 6 in front of O2.

Converting Grams to Moles

• Use molar mass to convert grams to moles:

  • Molar mass of C6H12O6:

    • C: 6x12.01 = 72.06 g,

    • H: 12x1.008 = 12.096 g,

    • O: 6x16.00 = 96.00 g,

    • Total = 72.06 + 12.096 + 96.00 = 180.2 g.

• To find moles:

  • Moles of C6H12O6 = 25 g / 180.2 g/mol = 0.1385 moles.

Moles to Grams Conversion

• To find grams of O2 produced from glucose:

  • Use mole ratio from the balanced equation: 1 mole C6H12O6 yields 6 moles O2.

  • Moles of O2 = 0.1385 moles C6H12O6 x (6 moles O2 / 1 mole C6H12O6) = 0.831 moles O2.

  • Convert to grams:

    • Molar mass of O2 = 32 g/mol.

    • Grams of O2 = 0.831 moles x 32 g/mol = 26.6 g O2.

Limiting Reactants

• In experiments, a limiting reactant is a reactant that is completely consumed in a reaction, limiting the amount of product produced.

• Example with bicycle construction:

  • Each bike requires:

    • 2 wheels,

    • 1 frame,

    • 1 handlebar.

  • With available materials:

    • 10 handlebars → 10 bikes,

    • 12 frames → 12 bikes,

    • 22 wheels → 11 bikes.

  • Limiting component is wheels; maximum bikes made is 10.

Real-Life Lab Considerations

• In real-life lab settings, yields can be less than predicted due to:

  • Spills or incorrect measurements,

  • One reactant running out before the other.

• Understanding the limiting reactant helps predict how much product is actually formed.

Example Reaction & Stoichiometry Problems

• Reaction of CO + O2 forms CO2:

  • If starting with 4 CO and 3 O2,

    • 4 CO can form 4 CO2,

    • 3 O2 can form up to 6 CO2 (since the ratio is 2:1),

    • Limiting reactant is CO, yielding 4 CO2;

    • O2 left over: 1 O2.

Steps to Identify Limiting Reactants

1. Write balanced chemical equation.

2. Calculate moles for each reactant.

3. Use mole ratios to determine how much product each reactant can produce.

4. The reactant producing the least amount is the limiting reactant.

Calculating Theoretical Yield

• Start with limiting reactant to calculate theoretical yield.

• For products, use balanced equation coefficients to find how much product can theoretically be generated.

• If reacting 12 moles of Ba(OH)2 with 5 moles of H3PO4:

  • Balanced Eq: 3Ba(OH)2 + 2H3PO4 → Ba3(PO4)2 + 6H2O.

  • From Ba(OH)2 (12 moles) → 4 moles Ba3(PO4)2.

  • From H3PO4 (5 moles) → 2.5 moles Ba3(PO4)2;

    • The limiting reactant is H3PO4.

Percent Yield Calculation

• Percent Yield = (Actual Yield / Theoretical Yield) x 100.

• Example: If produced 212 grams HCL when theoretical yield is 337.2 grams:

  • Percent Yield = (212/337.2) x 100 = 62.9%.

Conclusion

• To solve limiting reactant problems:

  • Regularly balance chemical equations, understand mole ratios, and practice convert between grams and moles.

  • Familiarity with each step builds confidence for tests and practical labs, including calculations for actual yield versus theoretical yield.