Problem 15 - Chemistry and Chemical Reactivity

Question

The reaction of methane and water is one way to prepare hydrogen for use as a fuel: $$ \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) $$ If you begin with \(995 \mathrm{g}\) of \(\mathrm{CH}_{4}\) and \(2510 \mathrm{g}\) of water, (a) Which reactant is the limiting reactant? (b) What is the maximum mass of \(\mathrm{H}_{2}\) that can be prepared? (c) What mass of the excess reactant remains when the reaction is completed?

Step-by-Step Solution

Answer

(a) Limiting reactant is CH4. (b) Maximum H2 mass is 375.91 g. (c) Excess H2O remaining is 1391.51 g.

Step 1: Convert Mass to Moles

First, convert the mass of each reactant to moles using their molar masses. The molar mass of methane (\(\mathrm{CH}_4\)) is 16.04 \(\mathrm{g/mol}\) and for water (\(\mathrm{H}_2 \mathrm{O}\)) it is 18.02 \(\mathrm{g/mol}\).\[\text{Moles of } \mathrm{CH}_4 = \frac{995 \, \mathrm{g}}{16.04 \, \mathrm{g/mol}} = 62.03 \, \text{moles}\]\[\text{Moles of } \mathrm{H}_2 \mathrm{O} = \frac{2510 \, \mathrm{g}}{18.02 \, \mathrm{g/mol}} = 139.28 \, \text{moles}\]

Step 2: Determine the Limiting Reactant

According to the balanced equation, 1 mole of \(\mathrm{CH}_4\) reacts with 1 mole of \(\mathrm{H}_2 \mathrm{O}\). Compare the mole ratios:\[62.03 \, \text{moles of } \mathrm{CH}_4 \quad ext{needs} \quad 62.03 \, \text{moles of } \mathrm{H}_2 \mathrm{O}\]Since there are 139.28 moles of \(\mathrm{H}_2 \mathrm{O}\), and only 62.03 moles of \(\mathrm{CH}_4\) are available, \(\mathrm{CH}_4\) is the limiting reactant.

Step 3: Calculate Maximum Hydrogen Produced

Using the limiting reactant \(\mathrm{CH}_4\), calculate the maximum moles of \(\mathrm{H}_2\) produced. According to the balanced equation,1 mole of \(\mathrm{CH}_4\) yields 3 moles of \(\mathrm{H}_2\):\[\text{Moles of } \mathrm{H}_2 = 62.03 \, \text{moles of } \mathrm{CH}_4 \times 3 = 186.09 \, \text{moles}\]Convert this to grams using the molar mass of \(\mathrm{H}_2\) (2.02 \(\mathrm{g/mol}\)):\[\text{Mass of } \mathrm{H}_2 = 186.09 \, \text{moles} \times 2.02 \, \mathrm{g/mol} = 375.91 \, \mathrm{g}\]

Step 4: Calculate Excess Water Remaining

Determine the amount of water that has reacted. Using the moles of the limiting reactant:\[\text{Moles of } \mathrm{H}_2 \mathrm{O} \text{ used} = 62.03 \, \text{moles}\]Subtract this from the initial moles of water:\[\text{Excess moles of } \mathrm{H}_2 \mathrm{O} = 139.28 - 62.03 = 77.25 \, \text{moles}\]Convert to grams:\[\text{Mass of excess } \mathrm{H}_2 \mathrm{O} = 77.25 \, \text{moles} \times 18.02 \, \mathrm{g/mol} = 1391.51 \, \mathrm{g}\]

Key Concepts

Mole Calculations

Mole calculations are foundational in chemistry for understanding the quantities of substances involved in chemical reactions. By determining the moles, you relate the masses of different compounds or elements to the number of particles (atoms, molecules, ions). To calculate moles, you need to use the molar mass, which is the mass of one mole of a substance, usually expressed in grams per mole. For example, consider methane (\(\mathrm{CH}_4\)) with a molar mass of 16.04 g/mol. If you have 995 grams of methane, you divide the mass by its molar mass: \[\text{Moles of } \mathrm{CH}_4 = \frac{995 \, \text{g}}{16.04 \, \text{g/mol}} = 62.03 \, \text{moles}\] Similarly, to find the moles of water (\(\mathrm{H}_2 \mathrm{O}\)) with a molar mass of 18.02 g/mol, you perform: \[\text{Moles of } \mathrm{H}_2 \mathrm{O} = \frac{2510 \, \text{g}}{18.02 \, \text{g/mol}} = 139.28 \, \text{moles}\] These calculations allow you to compare and use the quantities of reactants in the chemical reaction effectively.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of bonds, resulting in new products. Every chemical reaction requires reactants, and these reactants undergo changes as they convert into products following a balanced chemical equation. The balanced equation tells you the proportion of reactants required and the amount of products formed. For example, in the reaction: \(\mathrm{CH}_4(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{g})\), one mole of \(\mathrm{CH}_4\) reacts with one mole of \(\mathrm{H}_2 \mathrm{O}\) to produce one mole of \(\mathrm{CO}\) and three moles of \(\mathrm{H}_2\). This information forms the basis for calculating the limiting reactant and determining how much product can be formed from given amounts of reactants.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It involves using the balanced chemical equation to calculate the relative quantities of substances involved. An important concept of stoichiometry is identifying the limiting reactant—the substance that runs out first and thus limits the extent of the reaction. In our example reaction, we determined that methane (\(\mathrm{CH}_4\)) is the limiting reactant because its amount is less than the required ratio compared to water.Stoichiometry also allows us to calculate the theoretical yield of products. For instance, for every mole of methane, three moles of hydrogen are produced. Therefore, if you have 62.03 moles of methane, it should produce:\[\text{Moles of } \mathrm{H}_2 = 62.03 \, \text{moles} \times 3 = 186.09 \, \text{moles}\] To find the mass of this produced hydrogen, you multiply by the molar mass of hydrogen (2.02 g/mol):\[\text{Mass of } \mathrm{H}_2 = 186.09 \, \text{moles} \times 2.02 \, \text{g/mol} = 375.91 \, \text{g}\] Stoichiometry, therefore, allows chemists to predict how much product will result from the reactants used and vice versa.