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In a chemical reaction, not all reactants are completely used up. One of them will run out first and restrict how far the reaction can proceed. This reactant is known as the **limiting reactant**. Identifying the limiting reactant is crucial because it determines the maximum amount of products that can be formed. In our exercise, we have methane (CHâ‚„) and water (Hâ‚‚O) reacting. We first need to compare the mole amounts of CHâ‚„ and Hâ‚‚O using their molar masses.

• Molar mass of CHâ‚„: 16.05 g/mol
• Molar mass of Hâ‚‚O: 18.02 g/mol

Calculate moles for each:

• Moles of CHâ‚„: \( \frac{500}{16.05} \approx 31.15 \) mol
• Moles of Hâ‚‚O: \( \frac{1300}{18.02} \approx 72.15 \) mol

The balanced chemical equation shows a 1:1 ratio, meaning we need equal moles of CHâ‚„ and Hâ‚‚O. Since we have more moles of Hâ‚‚O than CHâ‚„, methane runs out first, making it the limiting reactant. Recognizing this lets us calculate how much Hâ‚‚ and the leftover reactant remain in reaction.

Moles serve as a bridge between the atomic scale and macroscopic measurements in stoichiometry problems. To understand the extent of chemical reactions, calculating the moles of reactants and products is vital. In stoichiometry, the amount of substance isn't directly measured in grams or liters but in moles. **Here's how we calculate the moles of reactants:**First, determine the molar mass for each reactant. Methane (CHâ‚„) has a molar mass of 16.05 g/mol, while water (Hâ‚‚O) is 18.02 g/mol. Using the mass provided in the exercise, we apply the formula:\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]For CHâ‚„, we calculate:\[ \frac{500 \text{ g}}{16.05 \text{ g/mol}} \approx 31.15 \text{ mol of CH}_4 \]And for Hâ‚‚O:\[ \frac{1300 \text{ g}}{18.02 \text{ g/mol}} \approx 72.15 \text{ mol of H}_2\text{O} \]These mole calculations are essential for determining the limiting reactant and predicting the amounts of products formed in the reaction.

Chemical reactions describe how substances transform into new products. For this, it is crucial to comprehend the balanced reaction equation.In this problem, we use the chemical equation:\[ \text{CH}_4 + \text{H}_2\text{O} \rightarrow \text{CO} + 3\text{H}_2 \]This equation shows that one molecule of methane reacts with one molecule of water to produce one molecule of carbon monoxide and three molecules of hydrogen gas. The coefficients tell us the **stoichiometric ratios** needed for balancing the reaction.Understanding the reaction mechanism is vital for anyone studying chemistry:

• The reactants are subtracted from the system, transforming into products.
• The products are generated based on the stoichiometric coefficients e.g., 3 moles of Hâ‚‚ produced per CHâ‚„ used.

In practical applications, reactions need specific percentages of components, especially in industries that manufacture chemicals. Stoichiometry guides these necessary calculations, ensuring efficiency and minimizing waste. By appreciating these reactions and their balance, students can foresee the outcomes of their chemical equations accurately.