6.3 Chemical Reactions and Chemical Equations

Chemical Reactions

• Chemical Change: A process where a substance undergoes a change and gets a new set of properties, losing its original characteristics.

• Chemical Reaction: The process through which chemical changes occur.

• Chemical Equation: A short way to represent a chemical reaction using symbols, formulas, and signs.

• Reactants: Substances that undergo the reaction.

• Products: New substances formed by the reaction.

Rules for Writing Chemical Equations:

1. Positioning of Reactants and Products:

• Reactants are written on the left side.

• Products are written on the right side.

• An equal sign (=) or an arrow (→) separates them.

2. Writing Symbols and Formulas:

• Use the symbols and formulas of the reactants and products.

• Multiple reactants or products are separated by a plus sign (+).

3. Balancing Equations:

• The number of atoms of each element on both sides should be equal. This ensures mass is conserved.

• Example: 2H2 + O2 → 2H2O

4. Arrow Usage:

• An arrow (→) is used instead of an equal sign (=) if the equation is unbalanced.

5. Physical States of Reactants and Products:

• Indicate physical states in brackets after the substance:

• (s) for solid

• (l) for liquid

• (g) for gas

• (aq) for aqueous solution (dissolved in water)

• Example: 2H2(g) + O2(g) → 2H2O(l)

6. Heat in Reactions:

• If heat is required, it is represented by a delta (Δ) symbol above the arrow.

• Example: 2Mg(NO3)2(s) → 2MgO(s) + 4NO2(g) + O2(g) (requires heat)

6.3.1 Balancing Chemical Equations

• Conservation of Mass: The number of atoms on both sides of the equation must be the same, meaning mass is conserved.

• Balancing Steps:

1. Write the correct chemical formula for reactants and products.

2. Balance compound substances first by adjusting their coefficients.

3. Balance elements atom by atom, ensuring equal numbers on both sides.

4. When all atoms are balanced, the equation is considered balanced.

Example 1:

• Unbalanced Reaction: Mg + HCl → MgCl2 + H2

• On the left: 1 Cl, 1 H, 1 Mg.

• On the right: 2 Cl, 2 H, 1 Mg.

• Balance Cl and H:

• Multiply HCl on the left by 2: Mg + 2HCl → MgCl2 + H2

• Now, both sides are balanced:

• Balanced Equation: Mg + 2HCl → MgCl2 + H2

Example 2:

• Unbalanced Reaction: Na2CO3 + HCl → NaCl + H2O + CO2

• Balance Na by multiplying NaCl by 2: Na2CO3 + HCl → 2NaCl + H2O + CO2.

• Balance Cl by multiplying HCl by 2: Na2CO3 + 2HCl → 2NaCl + H2O + CO2.

Example 3:

• Unbalanced Reaction: Al2O3 + HCl → AlCl3 + H2O

• Balance Al by multiplying AlCl3 by 2: Al2O3 + HCl → 2AlCl3 + H2O.

• Balance Cl by multiplying HCl by 6: Al2O3 + 6HCl → 2AlCl3 + H2O.

• Balance O and H by adjusting H2O: Al2O3 + 6HCl → 2AlCl3 + 3H2O.

6.3.2 Mole and Chemical Equation (Stoichiometry)

• Stoichiometry: The study of the quantitative relationships between reactants and products in a chemical reaction.

• It allows us to calculate how much of each substance is involved in a reaction.

Example Reaction:

• 2Mg(s) + O2(g) → 2MgO(s)

• 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO.

Stoichiometric Calculation:

• 2 moles of Mg = 48 grams.

• 1 mole of O2 = 32 grams.

• 2 moles of MgO = 80 grams.

Problem Example 1: How much oxygen is needed to react with 5 grams of magnesium?

• 48 grams of Mg react with 32 grams of O2.

• 1 gram of Mg reacts with 32/48 grams of O2.

• For 5 grams of Mg, the required O2 = (32/48) * 5 = 3.33 grams of O2.

Problem Example 2: How much magnesium oxide will be produced from 2 grams of magnesium?

• 48 grams of Mg produce 80 grams of MgO.

• 1 gram of Mg produces 80/48 grams of MgO.

• For 2 grams of Mg, the produced MgO = (80/48) * 2 = 3.33 grams of MgO.

Problem Example 3: How much oxygen is required to produce 10 grams of magnesium oxide?

• 80 grams of MgO are produced from 32 grams of O2.

• 1 gram of MgO is produced from 32/80 grams of O2.

• For 10 grams of MgO, the required O2 = (32/80) * 10 = 4 grams of O2.

Problem Example 4: How many NH3 molecules will be produced from 5 N2 molecules?

• From the equation: N2 + 3H2 → 2NH3.

• 1 molecule of N2 produces 2 molecules of NH3.

• For 5 molecules of N2, the number of NH3 molecules = 5 * 2 = 10 molecules of NH3.

Do It Yourself Tasks:

1. How many moles of O2 are necessary to produce 6 moles of water?

2. How much NH3 in liters will be formed from 4 liters of N2 at standard temperature and pressure (STP)?

These tasks help practice stoichiometry and balancing equations.