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

The atomic weight of metal is 55.8 (option c).

Step 1: Write the balanced chemical equation

The balanced chemical reaction for the reduction of metal oxide by hydrogen is: \( \mathrm{M}_{2} \mathrm{O}_{3} + 3\mathrm{H}_2 \rightarrow 2\mathrm{M} + 3\mathrm{H}_2\mathrm{O} \).This equation tells us that 3 moles of \( \mathrm{H}_2 \) react with 1 mole of \( \mathrm{M}_{2} \mathrm{O}_{3} \) to give 2 moles of \( \mathrm{M} \).

Step 2: Find moles of hydrogen used

We are given that 6 mg (0.006 g) of hydrogen are used. The molar mass of hydrogen \( \mathrm{H}_2 \) is 2 g/mol, so:\[ \text{Moles of } \mathrm{H}_2 = \frac{0.006}{2} = 0.003 \text{ moles} \]

Step 3: Relate moles of metal oxide to moles of hydrogen

From the balanced equation, 3 moles of \( \mathrm{H}_2 \) are needed to reduce 1 mole of \( \mathrm{M}_{2} \mathrm{O}_{3} \). Therefore, moles of \( \mathrm{M}_{2} \mathrm{O}_{3} \) that reacted are:\[ \text{Moles of } \mathrm{M}_{2} \mathrm{O}_{3} = \frac{0.003}{3} = 0.001 \text{ moles} \]

Step 4: Find molar mass of metal oxide

We have 0.1595 g of \( \mathrm{M}_{2} \mathrm{O}_{3} \) and it corresponds to 0.001 moles. Thus, the molar mass \( M_{M_2O_3} \) is:\[ M_{M_2O_3} = \frac{0.1595}{0.001} = 159.5 \text{ g/mol} \]

Step 5: Deduce metal atomic weight

Using the formula \( \mathrm{M}_{2} \mathrm{O}_{3} \), if \( M \) is the atomic weight of metal, the molar mass is given by:\[ 2M + 3 \times 16 = 159.5 \]Solving this: \[ 2M + 48 = 159.5 \]\[ 2M = 111.5 \]\[ M = \frac{111.5}{2} = 55.75 \approx 55.8 \]

Step 6: Select the correct answer

Comparing the calculated atomic weight of metal \( M = 55.8 \) with the given options, the correct answer is:(c) \( 55.8 \)

Key Concepts

Chemical Equations Balancing

Balancing chemical equations is essential in chemistry. It ensures that the same number of each type of atom appears on both sides of the equation, adhering to the law of conservation of mass. For this exercise, we're examining the reduction of a metal oxide by hydrogen. The balanced chemical equation is \[ \mathrm{M}_2 \mathrm{O}_3 + 3\mathrm{H}_{2} \rightarrow 2\mathrm{M} + 3\mathrm{H}_2\mathrm{O} \]This equation indicates that:

• 1 mole of \( \mathrm{M}_{2} \mathrm{O}_{3} \) reacts with 3 moles of \( \mathrm{H}_2 \).
• This reaction results in the formation of 2 moles of \( \mathrm{M} \) (metal), highlighting the stoichiometry of the reaction.

Understanding how to balance equations helps predict the amounts of reactants needed and products formed. It provides a foundation for deeper concepts like stoichiometry and mass calculations.

Moles and Molar Mass

The concept of moles allows chemists to count atoms and molecules in a manageable way. Understanding moles and molar mass is crucial for converting between mass and moles, which in turn is necessary for stoichiometry and chemical calculations.In this exercise, we calculate moles based on the provided masses:

• The molar mass of \( \mathrm{H}_2 \) is 2 g/mol.
• Given 6 mg (0.006 g) of hydrogen, we find the moles of \( \mathrm{H}_2 \) using the formula: \[ \text{Moles of } \mathrm{H}_2 = \frac{0.006}{2} = 0.003 \; \text{moles} \]
• This tells us how much hydrogen is involved in the reaction.

The understanding of moles and molar mass aids in determining precise quantitative relationships in a chemical reaction.

Atomic Weight Calculation

Atomic weight, also known as atomic mass, is the weight of an individual atom of a given element. It's usually expressed in atomic mass units (amu). This exercise involves calculating the atomic weight of a metal (\( M \)) in \( \mathrm{M}_2 \mathrm{O}_3 \).After determining the molar mass of the compound, we set up the equation based on the chemical formula:

• Molar mass of \( \mathrm{M}_{2} \mathrm{O}_{3} \) is 159.5 g/mol.
• The atomic weight calculation involves subtracting the mass of oxygen: \[ 2M + 3 \times 16 = 159.5 \]
• Solving gives \[ 2M = 159.5 - 48 = 111.5 \]
• Finally, dividing by 2 gives the atomic weight of the metal, \( M = 55.75 \approx 55.8 \).

Through calculations like these, we can deduce important characteristics of unknown metals.

Stoichiometry in Reduction Reactions

Stoichiometry is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It's particularly important in reduction reactions, where elements lose or gain electrons.In the reduction of \( \mathrm{M}_{2} \mathrm{O}_{3} \) by hydrogen, the stoichiometry is expressed in terms of moles:

• 3 moles of \( \mathrm{H}_2 \) are required to reduce 1 mole of \( \mathrm{M}_2 \mathrm{O}_{3} \).
• The reaction produces 2 moles of free metal \( \mathrm{M} \).
• This stoichiometric relation helps determine the amount of reactant needed and the amount of product formed.

By applying stoichiometry, we ensure that all components of the reaction are correctly accounted for, leading to more accurate calculations of chemical quantities.