Question

A mixture of \(20 \mathrm{ml}\) of methane and \(20 \mathrm{ml}\) of \(\mathrm{O}_{2}\) is exploded and cooled to room temperature. If the reaction between the two substances is written as $$ \mathrm{CH}_{4}+2 \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} $$ then the final volume of the gaseous mixture is (1) \(10 \mathrm{ml}\) (2) \(20 \mathrm{ml}\) (3) \(30 \mathrm{ml}\) (4) \(60 \mathrm{ml}\)

Short answer

The final volume of the gaseous mixture is 20 ml.

Step-by-step solution

- Write the Balanced Equation

Given the balanced chemical equation: \[\mathrm{CH}_{4}+2 \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\]

- Identify Initial Volumes

Initial volumes of the gases: Methane (\(\mathrm{CH}_{4}\)): 20 ml Oxygen (\(\mathrm{O}_{2}\)): 20 ml

- Determine Stoichiometric Ratio

From the reaction, 1 mole of \(\mathrm{CH}_{4}\) reacts with 2 moles of \(\mathrm{O}_{2}\). This means 20 ml of \(\mathrm{CH}_4\) will need 40 ml of \(\mathrm{O}_2\).

- Establish Limiting Reactant

Since there are only 20 ml of \(\mathrm{O}_2\) available, \(\mathrm{O}_{2}\) is the limiting reactant. This will determine how much \(\mathrm{CH}_4\) reacts.

- Calculate Volume of Reacted Methane and Oxygen

20 ml of \(\mathrm{O}_2\) will react with half of that volume of \(\mathrm{CH}_4\), which is 10 ml of \(\mathrm{CH}_4\).

- Evaluate Volume of Products Formed

Producing 1 mole of \(\mathrm{CO}_{2}\) from 1 mole of \(\mathrm{CH}_4\) means 10 ml of \(\mathrm{CH}_4\) will produce 10 ml of \(\mathrm{CO}_2\). Note that \(\mathrm{H}_2O\) is in liquid form at room temperature and does not contribute to the gas volume.

- Determine Final Volume of Gases Mixture

The final gaseous mixture will consist of: -Unreacted \(\mathrm{CH}_4\) : 10 ml -Produced \(\mathrm{CO}_2\) : 10 ml. Adding these, the final volume is 20 ml.

Key concepts

limiting reactant in chemical reactions

In any chemical reaction, the limiting reactant is the substance that runs out first and thus determines the amount of product(s) formed. Imagine you're baking cookies, but have only enough flour for one batch. Even if you have extra sugar and eggs, you're limited by that flour. In our methane and oxygen reaction, even though we started with 20 ml of both gases, the balanced chemical equation \( \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \) tells us that methane (CH4) requires twice the amount of oxygen (O2) to fully react. Since we have 20 ml of O2, it can only fully react with half that volume of methane - 10 ml. Hence, \( \text{O}_2 \) is the limiting reactant, controlling the reaction completion.

balanced chemical equations

A balanced chemical equation is crucial for understanding chemical reactions. It tells how much of each reactant is needed and how much of each product is formed. The balanced equation \( \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \), shows that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water. This is why for our 20 ml of CH4, we'd need 40 ml of O2 to react completely, but since we only had 20 ml of O2, it limited the reaction. Always ensure your equations are balanced to make accurate predictions in real-world reactions!

gas laws in chemical reactions

The gas laws help describe how gases behave under different conditions of temperature, volume, and pressure. In our reaction, we cooled the gases to room temperature after combustion. According to 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, at constant temperature and pressure, the volume of gas produced is directly proportional to the moles of reactants consumed. Since O2 is the limiting reactant, it dictates the reaction's extent. The 20 ml of O2 reacts fully with 10 ml of CH4, forming 10 ml of CO2, with water (produced) being liquid at room temperature, not contributing to the gas volume. The leftover CH4 (10 ml) and produced CO2 (10 ml) give our final gas volume of 20 ml.