Problem 120 - A Complete Resource Book in Chemistry for JEE Main

Question

A metal oxide has the formula \(\mathrm{M}_{2} \mathrm{O}_{3} .\) It can be reduced by hydrogen to give free metal and water. \(0.1595 \mathrm{~g}\) of the metal oxide requires \(6 \mathrm{mg}\) of hydrogen for complete reduction. What is the atomic weight of metal? (a) \(54.4\) (b) \(46.56\) (c) \(55.8\) (d) \(58.5\)

Step-by-Step Solution

Answer

Atomic weight of metal is 55.8.

Step 1: Determine Initial Reaction

The reduction of the metal oxide, \( \mathrm{M}_2\mathrm{O}_3 \), by hydrogen gas can be expressed as: \( \mathrm{M}_2\mathrm{O}_3 + 3\mathrm{H}_2 \rightarrow 2\mathrm{M} + 3\mathrm{H}_2\mathrm{O} \). In this balanced chemical equation, one mole of \( \mathrm{M}_2\mathrm{O}_3 \) reacts with 3 moles of \( \mathrm{H}_2 \) to produce 2 moles of metal \( \mathrm{M} \) and 3 moles of water.

Step 2: Calculate Moles of Hydrogen

Given that 6 mg of hydrogen is used, convert this to grams, which is \( 0.006 \) grams. The molar mass of hydrogen gas \( (\mathrm{H}_2) \) is 2 grams/mole. Thus, the number of moles of hydrogen is: \[ \text{moles of } \mathrm{H}_2 = \frac{0.006}{2} = 0.003 \text{ moles} \].

Key Concepts

Reduction Reaction

When we talk about reduction reactions, we are essentially discussing a process where a substance gains electrons. In chemistry, this usually involves a metal oxide losing its oxygen atoms to form the pure metal. Such reactions are crucial in the field of metallurgy and various industrial processes.
One classic example is the reduction of a metal oxide using hydrogen gas. In this type of reaction, hydrogen acts as a reducing agent, which essentially means it facilitates the removal of oxygen from the metal oxide. The balanced chemical equation gives us a clear picture of this process:

• The metal oxide (3M_23O_33) reacts with hydrogen (3H_23) to yield the pure metal (3M3) and water (3H_2O3).
• The equation demonstrates conservation of mass and atoms, as the number of each type of atom is the same on both sides of the reaction.

Understanding this concept is essential for comprehending how metals are extracted and purified in real-world applications.

Moles of Hydrogen

The concept of a mole is vital in understanding chemical reactions as it allows chemists to count atoms, molecules, or other particles by weighing them. To figure out the amount of substance used or produced in a reaction, we use the concept of moles.
When hydrogen is involved, we need to consider its molar mass. Hydrogen gas (3H_23) has a molar mass of 2 grams per mole. This means that if we have 2 grams of hydrogen gas, we have 1 mole of hydrogen molecules.
In the given exercise, 6 mg of hydrogen is used to reduce the metal oxide. Here's how you can calculate the moles of hydrogen used:

• First, convert the mass of hydrogen from milligrams to grams: 6 mg is equal to 0.006 grams.
• Next, use the molar mass of hydrogen gas to find the moles: The formula is \(\text{moles of } \mathrm{H}_2 = \frac{\text{mass of } \mathrm{H}_2}{\text{molar mass of } \mathrm{H}_2}\).
• Then, calculate: \[ \text{moles of } \mathrm{H}_2 = \frac{0.006}{2} = 0.003 \text{ moles} \]

So, there are 0.003 moles of hydrogen used in the reaction. This step is crucial because it directly relates to the stoichiometry of the reaction and helps determine the amount of metal produced.

Metal Oxide Reduction

Metal oxide reduction is a widely-used process in which a metal oxide is converted into a pure metal through the removal of its oxygen. This process is commonly performed using a reducing agent like hydrogen gas. In the exercise we are dealing with, the metal oxide has the formula 3M_23O_33, which means two metal atoms are bonded to three oxygen atoms.

• The goal is to convert 3M_23O_33 into pure metal 3M3.
• This process involves breaking the chemical bonds between metal and oxygen, then forming new bonds to produce free metal and water.

The reaction equation can help us visualize what's happening:\( \mathrm{M}_2\mathrm{O}_3 + 3\mathrm{H}_2 \rightarrow 2\mathrm{M} + 3\mathrm{H}_2\mathrm{O} \)
To think about it simply, the 3H_23 molecules join with the oxygen atoms, effectively "grabbing" them and leaving free metallic atoms behind. By understanding metal oxide reduction, students can see how raw materials like ores are transformed into useful metals, a process that is foundational to material science and engineering.