Question

Consider the combustion reaction hetween \(25.0 \mathrm{mI}\). 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.

Short answer

The limiting reactant in this reaction is oxygen gas, and when the reaction goes to completion, 0.744 moles of water will be formed.

Step-by-step solution

Write the balanced chemical equation

For the combustion of methanol, the balanced chemical equation is: \(2 \ \text{CH}_3\text{OH}(l) + 3 \ \text{O}_2(g) \rightarrow 2 \ \text{CO}_2(g) + 4 \ \text{H}_2\text{O}(g)\) Now we have the balanced chemical equation to proceed with the calculations.

Convert the volume of methanol to mass

We are given the volume of methanol and its density, which can be used to find the mass: Mass of methanol = Volume × Density Mass of methanol = \( 25.0 \ \text{mL} \times 0.850 \frac{\text{g}}{\text{mL}} = 21.25 \ \text{g} \)

Calculate the number of moles of methanol and oxygen

To find the number of moles, we will use the molar masses of methanol and oxygen: Moles of methanol = \(\frac{\text{mass of methanol}}{\text{molar mass of methanol}} = \frac{21.25 \ \text{g}}{32.04 \ \frac{\text{g}}{\text{mol}}} = 0.663 \ \text{mol}\) For oxygen gas, we are given the volume at STP. At STP, 1 mole of any gas occupies 22.4 L. Therefore: Moles of oxygen = \(\frac{\text{volume of oxygen}}{\text{molar volume at STP}} = \frac{12.5 \ \text{L}}{22.4 \ \frac{\text{L}}{\text{mol}}} = 0.558 \ \text{mol}\)

Determine the limiting reactant and calculate the number of moles of water formed

To determine the limiting reactant, we will compare the mole ratios of methanol and oxygen with the balanced chemical equation: Mole ratio of methanol to oxygen = \(\frac{0.663}{0.558} = 1.19\) From the balanced chemical equation, the required mole ratio of methanol to oxygen is: \( \frac{2 \ \text{moles of CH}_3\text{OH}}{3 \ \text{moles of O}_2} = \frac{2}{3} = 0.67\) Since the calculated mole ratio is greater than the required mole ratio (1.19 > 0.67), oxygen is the limiting reactant. Now, we can calculate the number of moles of water formed using the stoichiometry of the reaction: Moles of water formed = \( \frac{4 \ \text{moles of H}_2\text{O}}{3 \ \text{moles of O}_2} \times 0.558 \ \text{mol} \) Moles of water formed = 0.744 mol So, if the reaction goes to completion, 0.744 moles of water will be formed.

Key concepts

Understanding Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance combines with oxygen to produce heat and light. This is often seen when burning fuels like methanol. In the example above, methanol (\( ext{CH}_3 ext{OH}\)) reacts with oxygen (\( ext{O}_2\)) to produce carbon dioxide (\( ext{CO}_2\)) and water (\( ext{H}_2 ext{O}\)).
Some important points to remember:

• Combustion reactions are exothermic, meaning they release energy.
• The products typically include oxides; in organic compounds, these are often \( ext{CO}_2\) and \( ext{H}_2 ext{O}\).

This type of reaction is fundamental in both energy production and environmental studies.

Identifying the Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that runs out first, limiting the amount of products that can be formed. It can be determined by comparing the mole ratios from the balanced chemical equation.
To find the limiting reactant:

• Divide the moles of each reactant by their respective coefficients from the balanced equation.
• The reactant with the smallest resulting number is the limiting reactant.

In our example, oxygen is the limiting reactant, as its available mole ratio is less than what is required according to the equation. This determines the amount of \( ext{H}_2 ext{O}\) and \( ext{CO}_2\) produced.

Calculating Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (\( ext{g/mol}\)). It's important for converting between the mass of a substance and the number of moles.
To calculate molar mass:

• Sum the atomic masses of all atoms in a molecule.
• For methanol (\( ext{CH}_3 ext{OH}\)), the molar mass is 32.04 \( ext{g/mol}\).

Using molar mass, we converted 21.25 \( ext{g}\) of methanol to moles. This conversion is essential for stoichiometric calculations in any chemical reaction.

Balancing Chemical Equations

A balanced chemical equation ensures that the same number of each type of atom is present on both sides of the equation. This reflects the law of conservation of mass, stating that matter cannot be created or destroyed.
Steps to balance an equation:

• List each element and count the atoms on both sides.
• Add coefficients to balance the atoms for each element.
• Double-check your work to ensure all elements are balanced.

The balanced equation for methanol combustion is \(2 \, \text{CH}_3\text{OH}(l) + 3 \, \text{O}_2(g) \rightarrow 2 \, \text{CO}_2(g) + 4 \, \text{H}_2 ext{O}(g)\). Balancing is crucial as it dictates the proportions needed for the reaction to proceed smoothly.