Problem 119 - 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 volume of liquid \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion and you condense the water vapor.

Step-by-Step Solution

Answer

The volume of liquid water formed by condensing the water vapor produced in the combustion reaction between 25.0 mL of liquid methanol and 12.5 L of oxygen gas at STP is 13.4 mL.

Step 1: Write the balanced chemical equation

The combustion reaction of methanol (CH3OH) involves combining it with oxygen gas (O2) to form carbon dioxide (CO2) and water (H2O). First, write down the reactants and products: CH3OH + O2 → CO2 + H2O To balance the equation, we need to make sure the number of atoms for each element is the same on both sides: CH3OH + 3/2 O2 → CO2 + 2 H2O Now, the balanced chemical equation for the combustion of methanol is: 2 CH3OH + 3 O2 → 2 CO2 + 4 H2O

Step 2: Find the moles of reactants

To perform stoichiometry, we need the number of moles of each reactant. Given the volume and density of liquid methanol, we can find its mass and then its moles. As for the oxygen gas, its volume is given at STP (standard temperature and pressure), so we can use the molar volume of a gas at STP (22.4 L/mol) to convert the volume to moles: Moles of methanol: \(M_{CH3OH} = 25.0\,mL * 0.850\,g/mL = 21.25\,g\) \\ Molar mass of CH3OH = 32.04 g/mol \\ Moles of CH3OH: \(n_{CH3OH} = \frac{21.25\,g}{32.04\,g/mol} = 0.663\,mol\) Moles of oxygen: Molar volume of a gas at STP = 22.4 L/mol \\ Moles of O2: \(n_{O2} = \frac{12.5\,L}{22.4\,L/mol} = 0.558\,mol\)

Step 3: Determine the limiting reactant

Use stoichiometry to compare the moles of reactants and find which is limiting: Moles of O2 needed for complete reaction of CH3OH: \(0.663\,mol\,CH3OH * \frac{3\,mol\,O2}{2\,mol\,CH3OH} = 0.995\,mol\,O2\) Since we only have 0.558 mol of O2, oxygen is the limiting reactant.

Step 4: Calculate the moles of water formed

Use stoichiometry to find the number of moles of water formed, based on the limiting reactant (O2): Moles of H2O: \(0.558\,mol\,O2 * \frac{4\,mol\,H2O}{3\,mol\,O2} = 0.744\,mol\,H2O\)

Step 5: Convert moles of water vapor to volume of liquid water

To find the volume of liquid water, we can use the density of water (1.00 g/mL) and the molar mass of water (18.02 g/mol): Mass of H2O: \(0.744\,mol\,H2O * 18.02\,g/mol = 13.4\,g\,H2O\) \\ Volume of liquid H2O: \(V_{H2O} = \frac{13.4\,g}{1.00\,g/mL} = 13.4\,mL\) The volume of liquid water formed by condensing the water vapor produced in the reaction is 13.4 mL.

Key Concepts

Combustion Reaction

Combustion reactions are a type of chemical reaction where a substance combines with oxygen to release energy in the form of heat and light. In a typical combustion reaction, you start with a fuel and oxygen, and you end with carbon dioxide and water as products. These reactions are highly exothermic, meaning they release a lot of energy, which is why they're used in engines and heaters.

The combustion of methanol in the provided exercise is a prime example of such a reaction. Methanol ( CH_3OH ) reacts with oxygen ( O_2 ) to produce carbon dioxide ( CO_2 ) and water ( H_2O ). To ensure the reaction adheres to the law of conservation of mass, we balance the chemical equation. In our case, the balanced form is 2 CH_3OH + 3 O_2 → 2 CO_2 + 4 H_2O. This shows us the exact proportions of reactants and products, crucial for stoichiometric calculations.

Understanding combustion reactions not only helps in chemical calculations but also in real-world applications like energy production and pollution control.

Limiting Reactant

To understand limiting reactants, visualize making sandwiches. If you have eight slices of bread and only three slices of cheese, the cheese limits the number of sandwiches you can make. Similarly, in chemical reactions, the limiting reactant is the substance that is entirely consumed first, thus dictating the maximum amount of product that can be formed.

In the methanol combustion example, we calculated the moles of methanol ( 0.663 mol ) and oxygen ( 0.558 mol ). Given the stoichiometric coefficients from the balanced equation, we needed 0.995 mol of O_2 to fully react with all the CH_3OH . However, since we only have 0.558 mol of O_2 , the oxygen is the limiting reactant. It determines the extent of the reaction, which tells us how much water can be produced.

Identifying the limiting reactant is essential for optimizing chemical processes, ensuring maximum efficiency, and reducing waste.

Mole Calculations

Mole calculations are fundamental in chemistry, allowing us to convert between mass, volume, and the number of particles. The mole is a basic unit in chemistry that represents 6.022×10^{23} entities, often atoms or molecules. This concept is crucial when working with reactions, as it allows chemists to scale reactions from the molecular to the laboratory level.

In our exercise, we started with the volume and density of methanol and converted it to mass to find the number of moles. We did the same for oxygen gas using the molar volume at STP, which is 22.4 L/mol . These calculations help bridge the gap between the experimental data (like volume or mass) and the stoichiometric equations used to predict the amount of other substances involved or formed.

Mole calculations form the backbone of quantitative chemistry, ensuring reactions are accurate and products are predictable.

Density in Chemistry

Density is a property that relates the mass of a substance to its volume. It's an intrinsic property that can help identify substances and calculate their amounts in reactions. Density is expressed as mass per unit volume ( g/mL in liquids).

In the exercise, density helped convert the given volume of methanol ( 25.0 mL ) into mass. Knowing the mass and chemical formula of methanol, we derived the number of moles for further stoichiometric calculations. This conversion is essential when dealing with liquids, as their volume can vary with temperature and pressure, but their density provides a constant link to their mass.

Density not only helps in analytical calculations but also assists in practical lab settings where precise measurements are crucial for successful experiments and reproducible results.