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In a chemical reaction, it's essential to maintain the same number of each type of atom on both sides of the equation. This is done by balancing the chemical equation. For the combustion of acetylene (Câ‚‚Hâ‚‚), the balanced equation is:
\[2 \text{C}_2\text{H}_2 + 5 \text{O}_2 \rightarrow 4 \text{CO}_2 + 2 \text{H}_2\text{O} \text{ with the release of 2600 kJ of heat}.\]
Each side of the equation has the same number of carbon, hydrogen, and oxygen atoms:

• 4 Carbon atoms (2 molecules of Câ‚‚Hâ‚‚) → 4 Carbon atoms (4 molecules of COâ‚‚)
• 4 Hydrogen atoms (2 molecules of Câ‚‚Hâ‚‚) → 4 Hydrogen atoms (2 molecules of Hâ‚‚O)
• 10 Oxygen atoms (5 molecules of Oâ‚‚) → 10 Oxygen atoms (4 molecules of COâ‚‚ and 2 molecules of Hâ‚‚O)

This keeps the law of conservation of mass intact.

In an exothermic reaction, energy is released into the surroundings, usually in the form of heat. The combustion of acetylene (\text{C}_2\text{H}_2}) is an exothermic reaction because it releases 1300 kJ of heat per mole of acetylene. Since there are 2 moles of acetylene in the balanced equation, it releases 2600 kJ of heat:
\[2 \text{C}_2\text{H}_2 + 5 \text{O}_2 \rightarrow 4 \text{CO}_2 + 2 \text{H}_2\text{O} \text{ with the release of 2600 kJ of heat}.\]
Exothermic reactions are characterized by the release of heat, making the surrounding environment warmer. This type of reaction is the opposite of an endothermic reaction, which absorbs heat from the surroundings.

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows us to calculate the amount of products formed in a reaction based on the quantities of reactants used.
In the acetylene combustion problem, we use stoichiometry to calculate the moles of water produced and the grams of oxygen needed. From the balanced equation, we see that 5 moles of Oâ‚‚ react with 2 moles of Câ‚‚Hâ‚‚ to produce 2 moles of Hâ‚‚O.
To find how many moles of Hâ‚‚O are produced with 2 moles of Oâ‚‚: \[ \frac{2 \text{ moles } \text{H}_2\text{O}}{5 \text{ moles } \text{O}_2} \times 2.00 \text{ moles } \text{O}_2 = 0.8 \text{ moles } \text{H}_2\text{O} \] For the grams of Oâ‚‚ needed to react with 9.80 grams of Câ‚‚Hâ‚‚:

• Calculate moles of Câ‚‚Hâ‚‚: \[ \frac{9.80 \text{ g}}{26.04 \text{ g/mol}} = 0.376 \text{ moles of C}_2\text{H}_2 \]
• Moles of Oâ‚‚ needed: \[ \frac{5 \text{ moles } \text{O}_2}{2 \text{ moles } \text{C}_2\text{H}_2} \times 0.376 \text{ moles } \text{C}_2\text{H}_2 = 0.94 \text{ moles } \text{O}_2 \]
• Grams of Oâ‚‚ needed: \[ 0.94 \text{ moles } \text{O}_2 \times 32.00 \text{ g/mol} = 30.08 \text{ grams } \text{O}_2 \]

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is calculated by adding the atomic masses of all the atoms in a molecule. For instance, the molar mass of acetylene (Câ‚‚Hâ‚‚) is calculated as follows:

• Carbon (C) has an atomic mass of approximately 12.01 g/mol.
• Hydrogen (H) has an atomic mass of approximately 1.01 g/mol.

Each molecule of acetylene (Câ‚‚Hâ‚‚) has 2 carbon atoms and 2 hydrogen atoms:
\[ \text{Molar Mass of C}_2\text{H}_2 = 2 \times 12.01 \text{ g/mol} + 2 \times 1.01 \text{ g/mol} = 26.04 \text{ g/mol}. \] We use molar mass to convert between grams and moles. For example, to find the moles of acetylene in 9.80 grams:
\[ \frac{9.80 \text{ g}}{26.04 \text{ g/mol}} = 0.376 \text{ moles of C}_2\text{H}_2 \] Knowing molar mass helps in various calculations, letting us know the amount of substance present.