Problem 71 - Chemistry The Molecular Science

Question

This reaction can be used to generate hydrogen gas from methane: $$\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})$$ If you use \(500 . \mathrm{g} \mathrm{CH}_{4}\) and \(1300 . \mathrm{g}\) water: (a) Determine the limiting reactant. (b) Calculate the mass (g) of \(\mathrm{H}_{2}\) that can be produced. (c) Calculate the mass (g) of the excess reactant remaining when the reaction is complete.

Step-by-Step Solution

Answer

(a) Limiting reactant: CH₄. (b) Mass of H₂: 188.77 g. (c) Excess H₂O remaining: 739.82 g.

Step 1: Calculate Moles of Reactants

First, determine the molar mass of methane (CH₄) and water (H₂O). The molar mass of CH₄ is approximately 16.05 g/mol and for H₂O it's about 18.02 g/mol. Now calculate the moles of each reactant: \[ \text{Moles of CH}_4 = \frac{500 \, \text{g}}{16.05 \, \text{g/mol}} \approx 31.15 \, \text{mol} \]\[ \text{Moles of H}_2\text{O} = \frac{1300 \, \text{g}}{18.02 \, \text{g/mol}} \approx 72.15 \, \text{mol} \]

Step 2: Determine the Stoichiometry

Check the balanced chemical equation to find the stoichiometric ratio for the reactants. The reaction equation is:\[ \text{CH}_4 + \text{H}_2\text{O} \rightarrow \text{CO} + 3\text{H}_2 \]This shows a \(1:1\) molar ratio between CH₄ and H₂O.

Step 3: Identify the Limiting Reactant

Compare the mole ratio from the calculation with the stoichiometric ratio in the equation. Since the mole ratio should be \(1:1\), check if moles of CH₄ and H₂O are equal:Moles of CH₄ = 31.15 molMoles of H₂O = 72.15 molSince fewer moles of CH₄ are available than H₂O, CH₄ is the limiting reactant.

Step 4: Calculate Mass of Hydrogen Produced

Use the moles of the limiting reactant (CH₄) to calculate the amount of hydrogen (H₂) produced. From the balanced equation, 1 mole of CH₄ produces 3 moles of H₂.\[ \text{Moles of H}_2 = 31.15 \, \text{mol} \times 3 = 93.45 \, \text{mol} \] The molar mass of H₂ is approximately 2.02 g/mol. Thus, the mass of H₂ is:\[ \text{Mass of H}_2 = 93.45 \, \text{mol} \times 2.02 \, \text{g/mol} \approx 188.77 \, \text{g} \]

Step 5: Calculate Remaining Excess Reactant

Calculate how much H₂O is consumed by using the stoichiometric ratio with CH₄. Since CH₄ is limiting, it determines the reaction amount.Moles of H₂O used = Moles of CH₄ reacted = 31.15 molFind the remaining moles of H₂O:\[ \text{Remaining H}_2\text{O} = 72.15 \, \text{mol} - 31.15 \, \text{mol} = 41.00 \, \text{mol} \]Convert to mass:\[ \text{Mass of remaining H}_2\text{O} = 41.00 \, \text{mol} \times 18.02 \, \text{g/mol} \approx 739.82 \, \text{g} \]

Key Concepts

Limiting Reactant

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 (CH₄) and water (H₂O) reacting. We first need to compare the mole amounts of CH₄ and H₂O using their molar masses.

• Molar mass of CH₄: 16.05 g/mol
• Molar mass of H₂O: 18.02 g/mol

Calculate moles for each:

• Moles of CH₄: \( \frac{500}{16.05} \approx 31.15 \) mol
• Moles of H₂O: \( \frac{1300}{18.02} \approx 72.15 \) mol

The balanced chemical equation shows a 1:1 ratio, meaning we need equal moles of CH₄ and H₂O. Since we have more moles of H₂O than CH₄, methane runs out first, making it the limiting reactant. Recognizing this lets us calculate how much H₂ and the leftover reactant remain in reaction.

Moles Calculation

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 (CH₄) has a molar mass of 16.05 g/mol, while water (H₂O) 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 CH₄, we calculate:\[ \frac{500 \text{ g}}{16.05 \text{ g/mol}} \approx 31.15 \text{ mol of CH}_4 \]And for H₂O:\[ \frac{1300 \text{ g}}{18.02 \text{ g/mol}} \approx 72.15 \text{ mol of H}_2\text{O} \]These mole calculations are essential for determining the limiting reactant and predicting the amounts of products formed in the reaction.

Chemical 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 H₂ produced per CH₄ 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.