Chemical Reactions and Chemical Quantities: Balancing, Stoichiometry, Limiting Reagents, and Percent Yield

Ch.7 - Chemical Reactions and Chemical Quantities

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, ensuring that the number and type of atoms are conserved during a chemical reaction. This process reflects the law of conservation of mass.

• Key Principle: The number and type of atoms on both sides of the equation must be equal.

• Coefficients: The numbers placed in front of chemical formulas to indicate the number of units involved; these are called stoichiometric coefficients.

• Steps to Balance:

  1. List all elements present in the reactants and products.

  2. Count the number of atoms of each element on both sides.

  3. Adjust coefficients to balance each element, starting with the most complex molecule.

  4. If a fractional coefficient is used, multiply the entire equation by the denominator to obtain whole numbers.

• Example: Balance the combustion of butane: ___ C4H10 (g) + ___ O2 (g) → ___ H2O (l) + ___ CO2 (g) Balanced: 2 C4H10 (g) + 13 O2 (g) → 10 H2O (l) + 8 CO2 (g)

Practice Problems:

• Balance: ___ Al4C3 (s) + ___ H2O (l) → ___ Al2O3 (s) + ___ CH4 (g)

• Balance: ___ C2H5SH (g) + ___ O2 (g) → ___ CO2 (g) + ___ H2O (l) + ___ SO2 (g)

• Write the balanced equation for the combustion of methyl tert-butyl ether, C5H12O (l).

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a balanced chemical equation. It allows chemists to predict the amounts of substances consumed and produced in a reaction.

• Stoichiometric Calculations: Use the coefficients from the balanced equation to relate moles of reactants and products.

• Steps:

  1. Identify the given quantity and the unknown quantity.

  2. Convert the given quantity to moles (if necessary).

  3. Use the mole-to-mole ratio from the balanced equation to find moles of the unknown.

  4. Convert moles of the unknown to the desired units (grams, molecules, etc.).

• Example: How many grams of NO are produced when 15.0 g N2 reacts?

Practice Problems:

• How many moles of hydrazine are formed when 25.7 g ammonia reacts with excess nitrogen gas?

• If 6.92 × 1021 molecules of solid phosphorus react, how many grams of phosphorus hexafluoride are produced?

• Determine the density of 50.0 mL iron (II) bromide when 112.7 g iron (II) phosphate is produced.

Limiting Reagent

The limiting reagent is the reactant that is completely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed. The other reactant(s) are in excess.

• Theoretical Yield: The maximum amount of product that can be formed from the limiting reagent.

• Excess Reagent: The reactant(s) that remain after the reaction is complete.

• Steps to Identify Limiting Reagent:

  1. Convert all given quantities to moles.

  2. Use the balanced equation to determine the amount of product each reactant can produce.

  3. The reactant that produces the least amount of product is the limiting reagent.

  4. The other reactant(s) are in excess.

• Example: Cr2O3 (s) + 3 H2S (g) → Cr2S3 (s) + 3 H2O (l) What is the mass of Cr2S3 formed when 14.20 g Cr2O3 reacts with 12.80 g H2S?

Practice Problems:

• If 12.0 g C3H6, 10.0 g NH3, and 5.0 g O2 react, what mass of acrylonitrile (C3H3N) can be produced?

• If 150 g of Al are reacted with 432 g of Fe2O3, what is the mass of the excess reactant remaining?

Percent Yield

Percent yield measures the efficiency of a chemical reaction by comparing the actual yield to the theoretical yield.

• Percent Yield Formula: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

• Actual Yield: The amount of product actually obtained from a reaction (measured in the lab).

• Theoretical Yield: The maximum possible amount of product, calculated from the limiting reagent.

• Efficiency: The higher the percent yield, the more efficient the reaction. Typical values:

  • Excellent: >90%

  • Very Good: 80–90%

  • Good: 70–80%

  • Poor: <70%

• Note: Actual yield is always less than theoretical yield due to losses and side reactions.

• Example: 2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l) If 2.6 g C6H6 reacts with excess O2 to produce 1.25 g H2O, what is the percent yield of water?

Practice Problems:

• What is the percent yield for a reaction in which 22.1 g Cu is isolated by reacting 45.5 g Zn with 70.1 g CuSO4?

• How many grams of hydrazine are produced from 115.0 g NH3 if the reaction has an 81.5% yield?

• If the reduction of Fe2O3 only goes to 75% completion, how many moles of Fe2O3 are required to produce 0.850 moles of Fe?