Stoichiometry and Chemical Reactions: Mass Calculations in GOB Chemistry

Stoichiometry in Chemical Reactions

Introduction

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It allows chemists to predict the amounts of substances consumed and produced, based on balanced chemical equations and the law of conservation of mass.

Sodium and Oxygen Reaction

Balanced Chemical Equation

The reaction between sodium and oxygen produces sodium oxide: $4\text{Na}(s) + \text{O}_2(g) \rightarrow 2\text{Na}_2\text{O}(s)$

Key Concepts

• Mole Ratio: The coefficients in the balanced equation indicate the ratio in which reactants and products participate. For every 4 moles of Na, 1 mole of O2 is required to produce 2 moles of Na2O.

• Molar Mass: The mass of one mole of a substance, used to convert between grams and moles.

  • Molar mass of Na: 22.99 g/mol

  • Molar mass of O2: 32.00 g/mol

  • Molar mass of Na2O: (2 × 22.99) + 16.00 = 61.98 g/mol

• Stoichiometric Calculations: Use the mole ratio and molar masses to determine the mass of reactants or products.

Example Problems

• a. Calculating Mass of Product: To find the grams of Na2O produced from 57.5 g of Na:

  1. Convert grams of Na to moles: $\text{moles Na} = \frac{57.5\ \text{g}}{22.99\ \text{g/mol}}$

  2. Use the mole ratio (4 mol Na : 2 mol Na2O) to find moles of Na2O.

  3. Convert moles of Na2O to grams: $\text{grams Na}_2\text{O} = \text{moles Na}_2\text{O} \times 61.98\ \text{g/mol}$

• b. Calculating Mass of Reactant Needed: To find grams of O2 needed for 18.0 g of Na:

  1. Convert grams of Na to moles.

  2. Use the mole ratio (4 mol Na : 1 mol O2) to find moles of O2.

  3. Convert moles of O2 to grams: $\text{grams O}_2 = \text{moles O}_2 \times 32.00\ \text{g/mol}$

• c. Calculating Mass of O2 Needed for a Given Product: For 75.0 g of Na2O produced:

  1. Convert grams of Na2O to moles.

  2. Use the mole ratio (2 mol Na2O : 1 mol O2) to find moles of O2.

  3. Convert moles of O2 to grams.

Nitrogen and Hydrogen Reaction

Balanced Chemical Equation

The reaction between nitrogen gas and hydrogen gas produces ammonia: $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$

Key Concepts

• Mole Ratio: 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

• Molar Mass:

  • Molar mass of N2: 28.02 g/mol

  • Molar mass of H2: 2.02 g/mol

  • Molar mass of NH3: 17.03 g/mol

Example Problems

• a. Calculating Mass of Product: To find grams of NH3 produced from 3.64 g of H2:

  1. Convert grams of H2 to moles: $\text{moles H}_2 = \frac{3.64\ \text{g}}{2.02\ \text{g/mol}}$

  2. Use the mole ratio (3 mol H2 : 2 mol NH3) to find moles of NH3.

  3. Convert moles of NH3 to grams: $\text{grams NH}_3 = \text{moles NH}_3 \times 17.03\ \text{g/mol}$

• b. Calculating Mass of Reactant Needed: To find grams of H2 needed for 2.80 g of N2:

  1. Convert grams of N2 to moles.

  2. Use the mole ratio (1 mol N2 : 3 mol H2) to find moles of H2.

  3. Convert moles of H2 to grams.

• c. Calculating Mass of Product from Reactant: For 12.0 g of H2:

  1. Convert grams of H2 to moles.

  2. Use the mole ratio to find moles of NH3.

  3. Convert moles of NH3 to grams.

General Steps for Stoichiometric Calculations

1. Write the balanced chemical equation.

2. Convert given mass to moles using molar mass.

3. Use the mole ratio from the balanced equation to relate moles of given substance to moles of desired substance.

4. Convert moles of desired substance to grams using its molar mass.

Summary Table: Stoichiometric Relationships

Reaction	Mole Ratio	Molar Masses (g/mol)
4Na + O2 → 2Na2O	4 Na : 1 O2 : 2 Na2O	Na: 22.99, O2: 32.00, Na2O: 61.98
N2 + 3H2 → 2NH3	1 N2 : 3 H2 : 2 NH3	N2: 28.02, H2: 2.02, NH3: 17.03

Applications

• Stoichiometry is essential for predicting yields in chemical manufacturing, laboratory synthesis, and environmental chemistry.

• It ensures efficient use of resources and helps avoid excess waste.