Problem 14 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

The equation for the complete combustion of methane is $$\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ What volume of oxygen at SATP is needed to react exactly with \(10 g\) of methane? (Section 8.2 )

Step-by-Step Solution

Answer

30.9 liters of oxygen are needed at SATP.

Step 1: Calculate Molar Mass of Methane

First, find the molar mass of methane (\(\mathrm{CH}_4\)). Methane consists of one carbon atom and four hydrogen atoms. The atomic mass of carbon (C) is approximately 12.01 g/mol, and that of hydrogen (H) is approximately 1.01 g/mol. Thus, the molar mass of methane \(\mathrm{CH}_4\) is \(12.01 + 4 \times 1.01 = 16.05\) g/mol.

Step 2: Determine Moles of Methane

Now we need to find out how many moles of methane are present in 10 g. Use the relationship: \( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \). For 10 g of methane, the moles \( = \frac{10}{16.05} \approx 0.623\) moles.

Step 3: Use Stoichiometry to Find Moles of Oxygen

According to the balanced chemical equation \(\mathrm{CH}_4(\mathrm{g}) + 2 \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) + 2 \mathrm{H}_2\mathrm{O}(\mathrm{g})\), one mole of methane reacts with two moles of oxygen. Therefore, \(0.623\) moles of methane will require \(2 \times 0.623 = 1.246\) moles of oxygen.

Step 4: Calculate Volume of Oxygen at SATP

At Standard Ambient Temperature and Pressure (SATP, which is 25°C and 100 kPa), the molar volume of an ideal gas is 24.8 L/mol. To find the volume of oxygen needed, multiply the moles of oxygen by the molar volume: \(1.246 \times 24.8 = 30.9\) liters.

Key Concepts

Understanding Molar Mass

Molar mass is a fundamental chemical concept that plays a vital role in various calculations, including stoichiometry. It represents the mass of one mole of a substance, usually measured in grams per mole (g/mol). Calculating the molar mass involves summing the atomic masses of all atoms in a molecule. For instance, in the case of methane (\(\mathrm{CH}_4\)), we have

• One carbon atom with an atomic mass of about 12.01 g/mol
• Four hydrogen atoms, each with an atomic mass of about 1.01 g/mol

Thus, the molar mass of methane becomes \(12.01 + 4 \times 1.01 = 16.05\) g/mol.
This calculation is essential for converting between mass and moles when analyzing chemical reactions, allowing us to maintain a balanced equation effectively, which is critical in stoichiometry.

Significance of Chemical Reactions

A chemical reaction is a process where substances known as reactants transform into new substances termed products. In stoichiometry, understanding the relationship between reactants and products is key for determining how much product forms or how much reactant is needed.
For the complete combustion of methane, represented by the equation\[ \mathrm{CH}_4(\mathrm{g}) + 2 \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) + 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \]The coefficients in the balanced equation tell us the ratio of molecules involved in the reaction. Here, one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water. This balance allows chemists to predict the outcomes when varying amounts of reactants are introduced. Stoichiometry bridges these predictions through precise calculations, ensuring reactions proceed efficiently and correctly.

Exploring the Ideal Gas Law

The ideal gas law is a cornerstone of chemistry, incorporating aspects such as pressure, volume, temperature, and moles to describe the behavior of gases. Represented by the equation \( PV = nRT \), where

• \(P\) is pressure,
• \(V\) is volume,
• \(n\) is the number of moles,
• \(R\) is the universal gas constant,
• \(T\) is temperature in Kelvin.

In this exercise, the focus is on calculating gas volume under SATP conditions (Standard Ambient Temperature and Pressure). At SATP, the molar volume of an ideal gas is typically 24.8 L/mol.
Using the ideal gas law here simplifies how we determine that 1.246 moles of oxygen, required for complete methane combustion, occupy \(1.246 \times 24.8 = 30.9\) liters under these standardized conditions. Thus, the ideal gas law offers a convenient way to quantify gases involved in reactions.