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Chemical equations are symbolic representations of chemical reactions. They show the reactants, which are substances that start a reaction, and products, which are substances produced by the reaction. For example, the chemical equation for the reaction between nitrogen gas (\text{N}_2) and hydrogen gas (\text{H}_2) to produce ammonia (\text{NH}_3) is given as: \[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \] This equation tells us that one molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia.

The molar mass of a substance is the mass of one mole of its particles (atoms, molecules, or ions) and is usually expressed in grams per mole (\text{g/mol}). To find the molar mass of a compound, add the atomic masses of all the atoms in its molecular formula.

For example, the molar mass of \text{N}_2 is calculated as follows: \[ \text{Molar Mass of } \text{N}_2 = 2 \times 14.01 \text{ g/mol} = 28.02 \text{ g/mol} \] Similarly, molar masses for hydrogen gas (\text{H}_2) and ammonia (\text{NH}_3) are: \[ \text{Molar Mass of } \text{H}_2 = 2 \times 1.01 \text{ g/mol} = 2.02 \text{ g/mol} \] \[ \text{Molar Mass of } \text{NH}_3 = 1 \times 14.01 \text{ g/mol} + 3 \times 1.01 \text{ g/mol} = 17.03 \text{ g/mol} \] Knowing the molar masses is crucial for converting between grams and moles.

Mass-mole conversion involves converting the mass of a substance to the number of moles and vice versa, using the molar mass. This concept is essential in stoichiometry, which involves calculating the amounts of reactants and products in a chemical reaction. To convert grams to moles:

Use the formula: \[ \text{Moles} = \frac{\text{Mass (in grams)}}{\text{Molar Mass (in g/mol)}} \] For instance, given 3.64 g of \text{H}_2, the number of moles is: \[ \text{Moles of } \text{H}_2 = \frac{3.64 \text{ g}}{2.02 \text{ g/mol}} = 1.80 \text{ mol} \] To convert moles to grams:

Use the formula: \[ \text{Mass (in grams)} = \text{Moles} \times \text{Molar Mass (in g/mol)} \] For example, to find the mass of 1.20 moles of \text{NH}_3: \[ \text{Mass of } \text{NH}_3 = 1.20 \text{ mol} \times 17.03 \text{ g/mol} = 20.44 \text{ g} \] Such computations are vital for solving chemical reaction problems.

Balanced chemical equations have equal numbers of each type of atom on both sides of the equation. This balance reflects the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. Balancing equations ensures that you have the same amount of each element in the reactants and products.

For the given reaction: \[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \] Count the number of nitrogen and hydrogen atoms:

• Reactants: \[ \text{N}_2 = 2 \text{ N atoms}, \text{H}_2 = 3 \times 2 = 6 \text{ H atoms} \]
• Products: \[ \text{NH}_3 = 2 \times \text{NH}_3 = 2 \times (1 \text{ N atom} + 3 \text{ H atoms}) = 2 \text{ N atoms} + 6 \text{ H atoms} \]

The equation is balanced because it has 2 N atoms and 6 H atoms on both sides. Balanced equations allow us to use stoichiometric ratios for calculations.

In any chemical reaction, the starting substances are called reactants, while the substances formed are called products. For example, in the ammonia synthesis reaction: \[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]

• Reactants: \text{N}_2 (Nitrogen gas) and \text{H}_2 (Hydrogen gas)
• Products: \text{NH}_3 (Ammonia)

Understanding the roles of reactants and products helps in performing calculations and understanding chemical processes. Knowing which substances are consumed and which are produced is fundamental in stoichiometry and balancing equations.

In stoichiometry, we use the balanced equation to relate the amounts of reactants to the amounts of products. The coefficients in the equation also indicate the mole ratios (e.g., 1 mole of \text{N}_2 reacts with 3 moles of \text{H}_2 to produce 2 moles of \text{NH}_3).