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 \(\mathrm{gas}\) measured at STP. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) Calculate the volume of liquid \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion and you condense the water vapor.

Short answer

The volume of liquid H2O formed in the combustion reaction between 25.0 mL of liquid methanol and 12.5 L of oxygen gas, measured at STP, is approximately \(13.39~\text{mL}\).

Step-by-step solution

Balanced Equation for Combustion of Methanol

The balanced equation for the combustion of methanol (CH3OH) is: \[2~\text{CH}_{3} \text{OH} + 3~\text{O}_{2} \rightarrow 2~\text{CO}_{2} + 4~\text{H}_{2} \text{O}\]

Convert Volume of Methanol to Moles of Methanol

First, we'll convert the volume of methanol (25.0 mL) to grams using the given density: \(25.0~\text{mL}\) \(\times\) \(\frac{0.850~\text{g}}{1~\text{mL}}\) \(=\) \(21.25\text{g}\) Now, convert grams of methanol to moles using the molar mass (\(M_{\text{CH}_{3}\text{OH}} = 32.04~\text{g/mol}\)): \(\frac{21.25 ~\text{g}}{32.04 ~\text{g/mol}} = 0.663 ~\text{moles} ~\text{CH}_{3}\text{OH}\)

Convert Volume of Oxygen Gas to Moles of Oxygen Gas

To convert the volume of oxygen gas (12.5 L) to moles at STP (Standard Temperature and Pressure), use the molar volume (22.4 L/mol): \(\frac{12.5 ~\text{L}}{22.4 ~\text{L/mol}} = 0.558 ~\text{moles}~ \text{O}_{2}\)

Determine the Limiting Reactant

According to the balanced equation, the mole ratio of CH3OH to O2 is 2:3. Divide the moles of each reactant by its coefficient in the balanced equation: For methanol: \(\frac{0.663 ~\text{mol}~\text{CH}_{3}\text{OH}}{2} = 0.331\) For oxygen: \(\frac{0.558 ~\text{mol}~\text{O}_{2}}{3} = 0.186\) Since the value for oxygen is smaller, oxygen is the limiting reactant.

Calculate Moles of H2O Produced

Using the mole ratio from the balanced equation (4 moles H2O per 3 moles O2), calculate moles of H2O produced: \(0.558 ~\text{mol} ~\text{O}_{2}\) \(\times\) \(\frac{4 ~\text{mol} ~\text{H}_{2} \text{O}}{3 ~\text{mol} ~\text{O}_{2}}\) \(=\) \(0.744 ~\text{mol} ~\text{H}_{2}\text{O}\)

Convert Moles of H2O to Grams

Now, we can convert moles of H2O to grams using the molar mass (\(M_{\text{H}_{2}\text{O}} = 18.015 ~\text{g/mol}\)): \(0.744 ~\text{mol} ~\text{H}_{2}\text{O}\) \(\times\) \(\frac{18.015~\text{g}}{1~\text{mol}}\) \(=\) \(13.39\text{g }\text{H}_{2}\text{O}\)

Convert Grams of H2O to Volume in mL

Finally, we'll convert grams of H2O to volume in mL using the density of water (approximately 1 g/mL at normal conditions): \(13.39~\text{g}\) \(\times\) \(\frac{1~\text{mL}}{1~\text{g}}\) \(=\) \(13.39\text{mL }\text{H}_{2}\text{O}\) Thus, the volume of liquid H2O formed in this reaction is approximately \(13.39~\text{mL}\).

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a reaction. This involves calculations based on the balanced chemical equation and the mole concept. To demonstrate with our example of methanol combustion, once the chemical equation is balanced, we can see that 2 moles of CH₃OH react with 3 moles of O₂ to produce 2 moles of CO₂ and 4 moles of H₂O. Knowing this, we can calculate how much of each product is formed from a given amount of reactants. Stoichiometry is essential in ensuring that a reaction follows the law of conservation of mass, meaning that there is no loss or gain in mass after the chemical reaction.

Limiting Reactant

The concept of the limiting reactant is pivotal in stoichiometric calculations. It is the reactant that will be used up first in a chemical reaction, thereby determining the maximum amount of product that can be formed. In the combustion of methanol, we have two reactants: methanol and oxygen. By comparing the stoichiometric ratios and the moles of each reactant available, we identified oxygen as the limiting reactant. This is because it will exhaust first according to the balanced equation, limiting the formation of products. Recognizing the limiting reactant is crucial because it allows us to calculate the exact amounts of products that will be formed.

Mole-to-Mass Conversion

Converting between moles and grams is often necessary in stoichiometry and is done using the substance's molar mass. Mole-to-mass conversion bridges the gap between the molecular scale and the practical, measurable quantities. For methanol (CH₃OH), we used its molar mass of 32.04 g/mol to convert the mass of methanol to moles. Similarly, after finding the moles of water (H₂O) produced, we usd its molar mass of 18.015 g/mol to convert moles of water to grams. This conversion is essential for practical applications, such as knowing how much product can be physically obtained from a reaction.

Chemical Reaction Balancing

Balancing chemical reactions is a fundamental aspect of stoichiometry, ensuring that the same number of each type of atom appears on both sides of the equation, in accordance with the Law of Conservation of Mass. It sets the stage for all subsequent stoichiometric calculations. In the methanol combustion, the balanced chemical equation indicates the precise stoichiometric ratios needed for the reaction to occur without any excess of atoms. For every 2 molecules of methanol, 3 molecules of O₂ are required to yield 2 molecules of CO₂ and 4 molecules of H₂O. We rely on these ratios to progress through the steps of quantifying reactants and products.

Molar Volume at STP

Molar volume at Standard Temperature and Pressure (STP) is a conversion factor used to relate the volume of a gas to the number of moles at 0°C and 1 atmosphere of pressure. The molar volume of any ideal gas at STP is 22.4 liters per mole. This allows us to convert between the volume of a gas in liters and the amount of substance in moles. For instance, we converted the volume of oxygen gas measured at STP to moles using the molar volume, which is a crucial step in the process of determining the limiting reactant and the extent of the reaction. The concept of molar volume simplifies handling gases, since it's often easier to measure the volume of gas than to weigh it.