Problem 113 - Chemistry: The Central Science

Question

Consider the combustion reaction between \(25.0 \mathrm{~mL}\) of liquid methanol (density \(=0.850 \mathrm{~g} / \mathrm{mL}\) ) and \(12.5 \mathrm{~L}\) of oxygen gas measured at STP. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\). Calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion.

Step-by-Step Solution

Answer

The moles of H2O formed in the combustion reaction between methanol and oxygen gas to completion is 0.75 mol.

Step 1: Write down the balanced chemical equation

The balanced combustion reaction between methanol (CH3OH) and oxygen gas (​O2) is: CH3OH (l) + 1.5 O2 (g) → CO2 (g) + 2 H2O (g)

Step 2: Convert volume of methanol to moles

To convert volume to moles, first convert the volume of methanol to mass using the given density, then convert the mass to moles using the molar mass of methanol. Volume of methanol = 25.0 mL Density of methanol = 0.850 g/mL Mass of methanol = Volume × Density Mass of methanol = 25.0 mL × 0.850 g/mL = 21.25 g Molar mass of methanol (CH3OH) = 12.01g/mol (C) + 1.01g/mol (H) × 4 + 16g/mol (O) = 32.05 g/mol Moles of methanol = Mass / Molar Mass Moles of methanol = \( \frac{21.25 \mathrm{~g}}{32.05 \mathrm{~g} / \mathrm{mol}} \) = 0.66 mol

Step 3: Convert volume of oxygen gas to moles at STP

Use the given volume of oxygen gas under standard temperature and pressure (STP) to calculate the moles of oxygen gas. Standard molar volume at STP = 22.4 L/mol Volume of O2= 12.5 L Moles of O2 = Volume / Standard molar volume Moles of O2 = \( \frac{12.5 \mathrm{~L}}{22.4 \mathrm{~L} / \mathrm{mol}}\) = 0.56 mol

Step 4: Determine the limiting reactant

Use the stoichiometry from the balanced equation and the calculated moles of methanol and oxygen gas to find the limiting reactant. Methanol to Oxygen ratio = 1:1.5 (from balanced equation) Required moles of O2 for complete combustion = Moles of methanol × 1.5 Required moles of O2 = 0.66 mol × 1.5 = 0.99 mol As there are only 0.56 mol of O2 available, oxygen is the limiting reactant.

Step 5: Calculate moles of H2O produced

Since oxygen is the limiting reactant, use the stoichiometry from the balanced equation to calculate the moles of H2O produced. Oxygen to Water ratio = 1.5:2 (from balanced equation) Moles of H2O = Moles of O2 × \( \frac{2}{1.5} \) Moles of H2O = 0.56 mol × \( \frac{2}{1.5} \) = 0.75 mol Thus, 0.75 mol of H2O is formed if the combustion reaction goes to completion.

Key Concepts

Combustion Reaction

A combustion reaction is a chemical process where a substance combines with oxygen to release energy. In the case of methanol combustion, methanol (CH3OH) reacts with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). This reaction is exothermic, meaning it releases heat, which makes combustion reactions common in engines and power plants.
To understand combustion more deeply, one must realize that oxygen contributes significantly to the production of energy through emission of heat and light.
Some key features of combustion reactions include:

• The presence of oxygen as a reactant.
• The production of one or more oxygen-containing compounds.
• Release of heat and light as common outcomes.

A balanced chemical equation for a combustion reaction is essential to ensure that matter is conserved; all atoms on the reactant side must appear on the products side. This balance becomes crucial when calculating other quantities in stoichiometry, like the number of moles formed or reactants consumed.

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is entirely consumed first, limiting the amount of product that can form. This concept is pivotal in stoichiometry, as it dictates the maximum yield of a reaction.
In our methanol combustion exercise, oxygen is identified as the limiting reactant. Here, we calculated that the reaction needs 0.99 moles of oxygen to fully combust 0.66 moles of methanol. However, we only have 0.56 moles of oxygen available. Therefore, oxygen limits the reaction progress.

Determining the limiting reactant involves stoichiometric calculations:

• Calculate the amount of each reactant needed based on the balanced equation.
• Identify which reactant is available in smaller quantity relative to its required amount.
• Use the limiting reactant to calculate the maximum amount of products that can be formed.

Understanding the concept of a limiting reactant is crucial when predicting the quantities of products formed in any reaction. It ensures resources are efficiently used and can also help prevent unnecessary wastage of reactants.

Mole Calculation

Mole calculations are fundamental in stoichiometry and involve converting quantities between moles, mass, and volume. In the given exercise, we calculated the number of moles for both methanol and oxygen to understand the combustion reaction fully.
To convert volume into moles, we first needed to know the density of methanol and its molar mass. After finding the mass by multiplying by density, we divided by the molar mass to get moles.
Similarly, oxygen moles were calculated using the standard molar volume at STP (22.4 L/mol), which simplifies the process of converting volume to mole calculation.
Steps involved in mole calculation can include:

• Identifying the given information (volume, mass, or number of moles).
• Using conversion factors such as density or molar volume (at STP).
• Applying the relationship: Moles = Mass / Molar Mass for solids/liquids or Volume / Standard molar volume for gases.

Understanding these steps allows clear communication in both theoretical reactions and practical laboratory settings where precise measurements are essential.