91Ó°ÊÓ

A balanced chemical equation shows the relationship between reactants and products, displaying their transformation during a chemical reaction. Each substance in the equation is represented by its chemical formula, with numbers in front to indicate moles. These coefficients ensure that the atoms on the reactant side equal those on the product side, satisfying the Law of Conservation of Mass.

In our example, the balanced chemical equation is:

• 4 \( \mathrm{P}_4 \)(s) + 5 \( \mathrm{S}_8 \)(s) \( \rightarrow \) 4 \( \mathrm{P}_4 \mathrm{S}_{10} \)(s)

This equation shows that 4 moles of \( \mathrm{P}_4 \) react with 5 moles of \( \mathrm{S}_8 \) to form 4 moles of \( \mathrm{P}_4 \mathrm{S}_{10} \). When a chemical equation is balanced, it provides essential information for calculating reactant consumption and product formation.

The mole ratio is a crucial concept in chemistry, derived from the coefficients of a balanced chemical equation. It shows the proportional relationship among the substances involved in a chemical reaction. In our equation, the mole ratio is derived from the coefficients: 4:5:4. This means that for every 4 moles of \( \mathrm{P}_4 \) combined with 5 moles of \( \mathrm{S}_8 \), 4 moles of \( \mathrm{P}_4 \mathrm{S}_{10} \) are produced.

The mole ratio is useful for solving problems in stoichiometry and determining the limiting reactant. For instance, calculating the actual mole ratio with a given number of moles helps to identify which reactant runs out first, limiting the reaction progress. In our case, even if we start with equal moles of \( \mathrm{P}_4 \) and \( \mathrm{S}_8 \), the theoretical ratio of 0.8 tells us that less \( \mathrm{S}_8 \) is required per mole of \( \mathrm{P}_4 \), making \( \mathrm{S}_8 \) the limiting reactant.

Stoichiometry is the study of the quantitative relationships in chemical reactions. It uses balanced chemical equations to calculate the amounts of reactants and products. Stoichiometry interprets the mole ratios and helps determine how much product can be formed from specific quantities of reactants, considering any limiting reactant.

In our problem, stoichiometry is used to find out how many moles of \( \mathrm{P}_4 \mathrm{S}_{10} \) can be produced with given amounts of \( \mathrm{P}_4 \) and \( \mathrm{S}_8 \). By knowing that \( \mathrm{S}_8 \) is the limiting reactant, we apply the stoichiometric coefficients from \( \mathrm{S}_8 \) to \( \mathrm{P}_4 \mathrm{S}_{10} \) to calculate the product yield:

• 0.500 mol \( \mathrm{S}_8 \) \( \times \frac{4 \text{ mol } \mathrm{P}_4 \mathrm{S}_{10}}{5 \text{ mol } \mathrm{S}_8} = 0.400 \text{ mol } \mathrm{P}_4 \mathrm{S}_{10}\)

Using stoichiometry allows us to predict the outcomes of chemical reactions precisely.

Chemical yield refers to the amount of product formed in a chemical reaction. It is often expressed as the percentage of the theoretical yield, which is the maximum amount of product that could be formed under perfect conditions. However, in practice, the actual yield may be lower due to factors like incomplete reactions or side reactions.

In the context of this exercise, understanding the chemical yield involves analyzing the effect of altering reactant amounts. For instance, doubling \( \mathrm{P}_4 \) while maintaining \( \mathrm{S}_8 \) doesn't change the yield because \( \mathrm{S}_8 \) remains the limiting reactant. Doubling \( \mathrm{S}_8 \), however, shifts the limiting factor to \( \mathrm{P}_4 \), allowing the potential yield to increase to 0.500 mol of \( \mathrm{P}_4 \mathrm{S}_{10} \).

• Doubling \( \mathrm{P}_4 \): Yield remains at 0.400 mol \( \mathrm{P}_4 \mathrm{S}_{10} \).
• Doubling \( \mathrm{S}_8 \): Yield increases to 0.500 mol \( \mathrm{P}_4 \mathrm{S}_{10} \).

By studying chemical yield, you can make predictions on maximizing the efficiency of reactions.