Problem 120 - 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 \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). 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 after the combustion reaction between 25.0 mL of liquid methanol and 12.5 L of oxygen gas at STP, if the reaction goes to completion and water vapor is condensed, will be 13.4 mL.

Step 1: Calculate the moles of methanol

First, we need to determine the moles of methanol in the reaction. We can use the given density of methanol (0.850 g/mL) and the volume (25.0 mL) to determine the mass of methanol. Then we can use the molar mass of methanol (CH₃OH, 32.04 g/mol) to calculate the moles of methanol. Mass of methanol = volume × density Mass of methanol = 25.0 mL × 0.850 g/mL = 21.25 g Moles of methanol = mass / molar mass Moles of methanol = 21.25 g / 32.04 g/mol = 0.663 mol

Step 2: Calculate the moles of oxygen

Next, we need to determine the moles of oxygen gas in the reaction. We'll use the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. The gas is at STP (standard temperature and pressure), which means P = 1 atm and T = 273 K. Given: R = 0.0821 (L.atm)/(mol.K) Rearrange the ideal gas law to solve for moles (n): n = PV / (RT) Moles of oxygen = (1 atm × 12.5 L) / (0.0821 (L.atm)/(mol.K) × 273 K) Moles of oxygen = 0.558 mol

Step 3: Determine the limiting reactant

Now that we have the moles of both reactants, we need to determine which one is the limiting reactant. We'll use the balanced chemical equation: 2 CH₃OH(l) + 3 O₂(g) -> 2 CO₂(g) + 4 H₂O(g) Divide the moles of each reactant by the coefficients in the balanced equation to determine the limiting reactant: Methanol: 0.663 mol / 2 = 0.331 Oxygen: 0.558 mol / 3 = 0.186 Since oxygen has the lowest value, it's the limiting reactant.

Step 4: Calculate the moles of water produced

Now that we know the limiting reactant is oxygen, we can calculate the moles of water formed using stoichiometry. From the balanced equation, we can see that 4 moles of H₂O are produced for every 3 moles of O₂. Moles of water = (0.558 mol O₂) × (4 mol H₂O / 3 mol O₂) Moles of water = 0.744 mol

Step 5: Calculate the volume of liquid water formed

Finally, we'll use the moles of water formed and the density of water (1.00 g/mL) to calculate the volume of liquid water formed. First, we need to convert moles of water to mass: Mass of water = moles × molar mass Mass of water = 0.744 mol × 18.02 g/mol = 13.402 g Now we can use the density of water to calculate the volume of liquid water: Volume of water = mass / density Volume of water = 13.402 g / 1.00 g/mL = 13.4 mL This means that if the reaction goes to completion and the water vapor is condensed, 13.4 mL of liquid water will be formed.

Key Concepts

Limiting Reactant

In chemical reactions, reactants are combined to produce products. When the amounts of reactants are not in the exact proportions as required by the balanced chemical equation, one reactant will be consumed first, halting the reaction. This is known as the limiting reactant. To identify the limiting reactant, we compare the mole ratio of the reactants to the coefficients in the balanced equation. The reactant which, due to its smaller mole ratio, runs out first limits the amount of product that can be formed and is thus considered the limiting reactant.

In our example with methanol and oxygen, we divided the moles of each reactant by their respective coefficients in the balanced equation and found that oxygen was the limiting reactant because it had the lower value when compared to the stoichiometric ratio. Therefore, the oxygen dictated how much of the products, in this case water and carbon dioxide, could be formed.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure (P), volume (V), temperature (T), and number of moles (n) of an ideal gas. Expressed as PV=nRT, where R is the ideal gas constant, this law allows us to calculate one of these four variables when the other three are known.

In our exercise, we used the ideal gas law to calculate the moles of oxygen gas at standard temperature and pressure (STP). By knowing the conditions of STP, and with R provided, we were able to rearrange the ideal gas law to solve for the moles (n) of oxygen, which was a crucial step in our stoichiometric calculations.

Molar Mass

The molar mass of a substance is the mass of one mole of that substance. It is an essential concept in stoichiometry because it links the mass of a sample to the number of moles, a fundamental count in chemical reactions. The molar mass is usually given in grams per mole (g/mol) and can be calculated by summing the atomic masses of all the atoms in a molecule.

In the solution, we used the molar mass of methanol (32.04 g/mol) and water (18.02 g/mol) to convert between mass and moles. This conversion was important both for determining the limiting reactant and for calculating the final volume of liquid water produced.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It is built around the balanced chemical equation, which provides the mole ratios needed to convert between moles of reactants and moles of products. Understanding stoichiometry is critical for predicting the amounts of products formed and reactants needed.

In the given problem, stoichiometry was used to calculate the moles of water produced based on the limiting reactant, oxygen. This process required using the mole ratio between oxygen and water from the balanced equation (3 mol O₂: 4 mol H₂O). The result gave us the amount of water, in moles, which we then converted to volume, demonstrating how stoichiometry guides us from reactants to products in a clear, step-by-step manner.