Question

Consider the reaction between \(50.0 \mathrm{~mL}\) liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}\) (density \(=0.850 \mathrm{~g} / \mathrm{mL}\) ), and \(22.8 \mathrm{~L} \mathrm{O}_{2}\) at \(27^{\circ} \mathrm{C}\) and \(\mathrm{a}\) pressure of \(2.00\) atm. 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 number of moles of water formed if the reaction goes to completion is \(0.819\text{ mol}\).

Step-by-step solution

1. Write the balanced chemical equation for the reaction

The balanced chemical equation for the combustion of methanol is: \(\mathrm{CH}_{3}\mathrm{OH}(l)+\frac{3}{2}\mathrm{O}_{2}(g)\longrightarrow \mathrm{CO}_{2}(g)+2\mathrm{H}_{2}\mathrm{O}(g)\)

2. Convert the volume of liquid methanol to moles

First, we will find the mass of methanol: Mass of methanol = Volume x Density Mass of methanol = \(50.0\text{ mL}\) x \(0.850\frac{\text{g}}{\text{mL}}=42.5\text{ g}\) Now we will find the moles of methanol: Moles = Mass/Molar mass The molar mass of methanol (\(\mathrm{CH}_3\mathrm{OH}\)) = \(12.01 + 3\times1.01+16.00 = 32.05\text{ g/mol}\) Moles of methanol =\(\frac{42.5}{32.05}\)= \(1.326\text{ mol}\)

3. Determine the moles of oxygen gas by using the ideal gas law

First, we need to convert the temperature to Kelvin: Temperature = \((27+273.15) \text{ K} = 300.15\text{ K}\) Now, for oxygen, use the ideal gas law: \(PV=nRT\) Where: P = pressure (2.00 atm) V = volume (22.8 L) n = moles of O2 R = gas constant (\(0.0821\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\)) T = temperature (300.15 K) Solve for n: \(n=\frac{PV}{RT}=\frac{2.00\text{ atm}\times22.8\text{ L}}{0.0821\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\times300.15\text{ K}}=1.844\text{ mol}\)

4. Determine the limiting reactant

To find the limiting reactant, compare the moles of methanol and oxygen to their stoichiometric ratios: From the balanced equation, 1 mol of \(\mathrm{CH}_3\mathrm{OH}\) reacts with \(\frac{3}{2}\) mol of \(\mathrm{O}_2\). Divide the available moles by their stoichiometric ratios: For methanol: \(1.326 \div 1 = 1.326\) For oxygen: \(1.844 \div \frac{3}{2} = 1.229\) Since 1.229 is smaller than 1.326, Oxygen is the limiting reactant.

5. Calculate the moles of water formed by using the stoichiometry of the balanced chemical equation

From the balanced equation, 2 moles of water are produced for every \(\frac{3}{2}\) moles of oxygen reacted: Moles of water = \(1.229\times\frac{2}{\frac{3}{2}}=1.229\times\frac{4}{3}=0.819\text{ mol}\) Thus, the number of moles of water formed if the reaction goes to completion is \(0.819\text{ mol}\).

Key concepts

Chemical Reaction

A chemical reaction is a process where reactants are transformed into products through the breaking and forming of chemical bonds. It involves the rearrangement of atoms and is commonly represented by a chemical equation. In this exercise, we explored the chemical reaction where methanol (CH₃OH) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O), known as a combustion reaction.

Understanding how different substances interact during a chemical reaction allows us to predict the products and calculate the quantities of each substance involved. This process is crucial for fields such as chemistry, engineering, and environmental science, where precise measurements can be the difference between success and disaster.

Balanced Chemical Equation

A balanced chemical equation is essential for accurately describing a chemical reaction. It ensures that the law of conservation of mass is respected, meaning the number of atoms of each element is the same on both sides of the equation. In our textbook exercise, the balanced equation is: CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2 H₂O(g).

Balancing an equation requires finding the correct stoichiometric coefficients that make the number of atoms of each element equal on both reactant and product sides. This process is critical as it determines how much of each reactant is needed to create a certain amount of product, which is what stoichiometry is all about.

Ideal Gas Law

The ideal gas law is a fundamental equation relating the pressure, volume, temperature, and amount (in moles) of an ideal gas. It is usually expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin.

In this exercise, we used the ideal gas law to find the number of moles of oxygen gas present under specific conditions (27°C and 2.00 atm). By manipulating the ideal gas equation, we can solve for the desired variable when the other three are known. In practical applications, the ideal gas law allows scientists and engineers to predict how a gas will behave under changing conditions, which is vital in many industrial and laboratory settings.