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 approximately 55.8.

Step 1: Write the Reduction Reaction

The reduction reaction of the metal oxide with hydrogen can be expressed as: \[ \mathrm{Z}_{2}\mathrm{O}_{3} + 3\mathrm{H}_2 \rightarrow 2\mathrm{Z} + 3\mathrm{H}_2\mathrm{O} \] This equation shows that 1 mole of \( \mathrm{Z}_{2}\mathrm{O}_{3} \) reacts with 3 moles of hydrogen gas to produce 2 moles of free metal and water.

Step 2: Calculate the Moles of Hydrogen Used

The mass of hydrogen used in the reaction is given as \( 6 \mathrm{mg} \), which is equal to \( 0.006 \mathrm{~g} \). The molar mass of hydrogen is \( 2.016 \mathrm{~g/mol} \). Therefore, the moles of hydrogen used is: \[ \text{Moles of } \mathrm{H}_2 = \frac{0.006}{2.016} \approx 0.00298 \text{ moles} \]

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

From the balanced equation, 3 moles of hydrogen are required for 1 mole of \( \mathrm{Z}_{2}\mathrm{O}_{3} \). Therefore, the moles of \( \mathrm{Z}_{2}\mathrm{O}_{3} \) is: \[ \text{Moles of } \mathrm{Z}_{2}\mathrm{O}_{3} = \frac{0.00298}{3} \approx 0.0009933 \text{ moles} \]

Step 4: Calculate the Molar Mass of Metal Oxide

The mass of \( \mathrm{Z}_{2}\mathrm{O}_{3} \) is given as \( 0.1596 \mathrm{~g} \). Using the formula \( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \), the molar mass of \( \mathrm{Z}_{2}\mathrm{O}_{3} \) is: \[ \text{Molar Mass of } \mathrm{Z}_{2}\mathrm{O}_{3} = \frac{0.1596}{0.0009933} \approx 160.7 \mathrm{~g/mol} \]

Step 5: Calculate the Atomic Weight of the Metal

The molar mass of \( \mathrm{Z}_{2}\mathrm{O}_{3} \) is made up of 2 moles of metal (\(2Z\)) and 3 moles of oxygen (\(3\times16.00 = 48\mathrm{~g/mol}\)). Thus, the mass of 2 moles of \(Z\) is: \[ 160.7 - 48 = 112.7 \mathrm{~g/mol} \] So, the atomic mass of the metal \(Z\) is: \[ \frac{112.7}{2} = 56.35 \mathrm{~g/mol} \] The closest option is (a) \( 55.8 \).

Key Concepts

Metal Oxides

Metal oxides are compounds consisting of a metal combined with oxygen. These compounds are prevalent in nature and are often found as minerals. In a metal oxide such as \( \mathrm{Z}_{2} \mathrm{O}_{3} \), the subscript numbers represent the ratio of metal atoms to oxygen atoms. Here, for every two atoms of the metal \( Z \), there are three oxygen atoms. This specific stoichiometric ratio is crucial for various chemical reactions, particularly reduction reactions where metal oxides are converted back to pure metals.

Metal oxides have significant roles in both industrial and environmental chemistry. They are commonly used in processes like metallurgy to extract metals. Understanding the stoichiometry and composition of metal oxides helps in predicting how much of each reactant will be needed and the amount of product generated in a reaction.

Reduction Reactions

Reduction reactions involve the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion. In the context of metal oxides, reduction reactions are processes where a metal oxide is converted to its pure metal form. Let's take the reaction of \( \mathrm{Z}_{2} \mathrm{O}_{3} \) with hydrogen:
\[ \mathrm{Z}_{2}\mathrm{O}_{3} + 3\mathrm{H}_2 \rightarrow 2\mathrm{Z} + 3\mathrm{H}_2\mathrm{O}\]
In this reaction, hydrogen acts as the reducing agent. It donates electrons to the oxide ions in \( \mathrm{Z}_{2} \mathrm{O}_{3} \), forming water and leaving behind the elemental metal. This process is a common method in metallurgy to extract metals from their ores.

Understanding reduction involves recognizing the changes in oxidation states during the reaction. For the metal in \( \mathrm{Z}_{2} \mathrm{O}_{3} \), the oxidation state decreases as it moves from being bonded with oxygen to becoming a free metal.

Molar Mass Calculations

Molar mass calculations are fundamental in stoichiometry as they allow for the conversion between the mass of a substance and its amount in moles. This is particularly important in determining the proportions of reactants and products in a chemical reaction. For example, consider the calculation of the molar mass of \( \mathrm{Z}_{2} \mathrm{O}_{3} \):

The formula states that the compound consists of two metal atoms \( \mathrm{Z} \) and three oxygen atoms. We calculated the molar mass of oxygen as \( 3 \times 16.00 = 48 \mathrm{~g/mol} \). Deducting the mass of oxygen from the total molar mass \( 160.7 \mathrm{~g/mol} \) gives the combined mass of the metal in \( \mathrm{Z}_{2} \mathrm{O}_{3} \):

\[ 112.7 \mathrm{~g/mol}\]

Since this is for two atoms of \( \mathrm{Z} \), the molar mass of one atom of \( Z \) is:
\[ \frac{112.7}{2} = 56.35 \mathrm{~g/mol}\]
Molar mass calculations enable chemists to connect these values with actual measurements in a lab, ensuring accurate prediction and calculation of results.