Problem 124 - A Complete Resource Book in Chemistry for JEE Main

Question

A metal oxide has the formula \(\mathrm{Z}_{2} \mathrm{O}_{3} .\) It can be reduced by hydrogen to give free metal and water. \(0.1596 \mathrm{~g}\) of the metal oxide requires \(6 \mathrm{mg}\) of hydrogen for complete reduction. The atomic weight of the metal is (a) \(55.8\) (b) \(65.8\) (c) \(6.58\) (d) \(15.9\)

Step-by-Step Solution

Answer

The atomic weight of the metal is 55.8; option (a) is correct.

Step 1: Write the Reduction Reaction

The given metal oxide is reduced by hydrogen to form the free metal and water. The reaction can be written as: \[ \mathrm{Z}_2\mathrm{O}_3 + 3\mathrm{H}_2 \rightarrow 2\mathrm{Z} + 3\mathrm{H}_2\mathrm{O} \]This indicates that one mole of \(\mathrm{Z}_2\mathrm{O}_3\) reacts with 3 moles of \(\mathrm{H}_2\) to produce 2 moles of metal \(\mathrm{Z}\) and 3 moles of water.

Step 2: Calculate moles of Hydrogen Used

Given that 6 mg of hydrogen is used and the molar mass of hydrogen \(\mathrm{H}_2\) is 2 g/mol, we calculate the moles of hydrogen used:\[\text{Moles of } \mathrm{H}_2 = \frac{6 \times 10^{-3} \text{ g}}{2 \text{ g/mol}} = 3 \times 10^{-3} \text{ moles}\]

Step 3: Relate Moles of Hydrogen to Moles of Metal Oxide

According to the reaction equation, 3 moles of hydrogen reduce 1 mole of \(\mathrm{Z}_2\mathrm{O}_3\). So moles of \(\mathrm{Z}_2\mathrm{O}_3\) are:\[\text{Moles of } \mathrm{Z}_2\mathrm{O}_3 = \frac{3 \times 10^{-3} \text{ moles of } \mathrm{H}_2}{3} = 1 \times 10^{-3} \text{ moles}\]

Step 4: Calculate Molar Mass of the Metal Oxide

Given mass of \(\mathrm{Z}_2\mathrm{O}_3\) is 0.1596 g and moles are \(1 \times 10^{-3}\):\[\text{Molar mass of } \mathrm{Z}_2\mathrm{O}_3 = \frac{0.1596 \text{ g}}{1 \times 10^{-3} \text{ moles}} = 159.6 \text{ g/mol}\]

Step 5: Create Expression for Molar Mass Based on Atomic Weights

The molar mass of \(\mathrm{Z}_2\mathrm{O}_3\) can also be calculated from the formula: \[2M + 3 \times 16 = 159.6 \]This is derived from the molar mass of the metal \(2M\) and the oxygen \(3 \times 16 = 48\).

Step 6: Solve for Atomic Weight of Metal

Plug the known values into the equation from the previous step:\[2M + 48 = 159.6\]\[2M = 159.6 - 48 = 111.6\]\[M = \frac{111.6}{2} = 55.8\]

Step 7: Determine Correct Answer

From the calculation, the atomic weight of the metal \(\mathrm{Z}\) is 55.8, hence the correct option is (a) 55.8.

Key Concepts

Reduction Reactions

Reduction reactions are a fundamental type of chemical reaction where a substance gains electrons. In the case of metal oxides, reduction often involves the removal of oxygen to revert to its elemental form, using a reducing agent such as hydrogen.

Here, the metal oxide \(\mathrm{Z}_2\mathrm{O}_3\) is reduced by hydrogen as shown in the reaction: \[ \mathrm{Z}_2\mathrm{O}_3 + 3\mathrm{H}_2 \rightarrow 2\mathrm{Z} + 3\mathrm{H}_2\mathrm{O} \] In this process, hydrogen is oxidized, meaning it gains oxygen to form water, while the metal loses oxygen, indicating it is reduced.

• Reduction: Loss of oxygen from a substance.
• Oxidation: Gain of oxygen by a substance.

This interplay of reduction and oxidation is a vital aspect of redox reactions in chemistry.

Mole Concept

The mole concept in chemistry helps quantify substances at the molecular level. A mole indicates a specific quantity, akin to a dozen, and is used to express amounts of a chemical substance. The magic number associated with a mole is Avogadro's number, \(6.022 \times 10^{23}\) entities (atoms, molecules, etc.).

In our exercise, hydrogen is given in milligrams, which we convert to moles using the molar mass. The equation used is: \[\text{Moles of } \mathrm{H}_2 = \frac{\text{mass in grams}}{\text{molar mass}} = \frac{6 \times 10^{-3}}{2} = 3 \times 10^{-3} \text{ moles} \]This calculation is pivotal as it allows us to relate grams to moles, offering a bridge between the macroscopic and atomic worlds we examine in stoichiometry and chemical reactions.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction.
It helps us predict how much of a substance is needed or formed in a reaction.

In the reduction of \(\mathrm{Z}_2\mathrm{O}_3\), stoichiometry dictates that the reaction between one mole of \(\mathrm{Z}_2\mathrm{O}_3\) and three moles of hydrogen produces two moles of the free metal, \ \mathrm{Z}\.

This stoichiometric ratio becomes a useful tool:

• 1 mole of \(\mathrm{Z}_2\mathrm{O}_3\) reacts with 3 moles of \(\mathrm{H}_2\)
• Produces 2 moles of \(\mathrm{Z}\)

Stoichiometry, therefore, enables us to deduce quantities of reactants and products involved, ensuring balanced and efficient chemical processes.

Metal Oxides

Metal oxides are compounds consisting of metal atoms bonded to oxygen atoms. They form part of many minerals and materials, important both industrially and biologically.

In the exercise, \(\mathrm{Z}_2\mathrm{O}_3\) represents a generic formula for a metal oxide, illustrating how metals can bind to oxygen.
This compound is reduced to their elemental metal, demonstrating the relationship during reduction reactions.

Metal oxides often behave as bases in reactions, reacting with acids to form salt and water. Their reactions such as reduction are essential in metallurgy for extracting pure metals from their oxides, which is critical in various industries.

Chemical Calculations

Chemical calculations translate a chemical formula or equation into numerical data, aiding in problem-solving involving moles, mass, or volume.

For \(\mathrm{Z}_2\mathrm{O}_3\) reduction, calculations verify the atomic weight of the metal involved. Using the known mass and moles, we find the molar mass.

Two approaches to verify the atomic weight can be done:

• Mass-based calculations
• Equation formulation: \ 2M + 48 = 159.6 \ (with \(48\) being the oxygen component for three atoms)

Simplifying and solving provides the atomic weight of the metal (M), showing practical use of chemical calculations in determining unknowns in chemical substances.