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 the metal is 55.8, which corresponds to option (c).

Step 1: Write the Reduction Reaction

The given metal oxide is \( M_2O_3 \). The reduction reaction can be written as: \[ M_2O_3 + 3H_2 \rightarrow 2M + 3H_2O \]

Step 2: Calculate Moles of Hydrogen Used

Convert the given mass of hydrogen into moles. \(\text{Mass of } H_2 = 6 \text{ mg} = 0.006 \text{ g} \). The molar mass of \(H_2\) is approximately \(2 \text{ g/mol}\). Thus, the moles of \(H_2\) is: \[ \text{Moles of } H_2 = \frac{0.006}{2} = 0.003 \text{ mol} \]

Step 3: Calculate Moles of Metal Oxide Used

Determine the moles of \(M_2O_3\). The reduction reaction shows 3 moles of \(H_2\) react with 1 mole of \(M_2O_3\). Thus, \(0.003\) moles of \(H_2\) will react with: \[ \text{Moles of } M_2O_3 = \frac{0.003}{3} = 0.001 \text{ mol} \]

Step 4: Determine the Molar Mass of Metal Oxide

Calculate the molar mass of \(M_2O_3\). \(0.1595 \text{ g} \) of \(M_2O_3\) corresponds to \(0.001 \text{ mol}\). Thus, the molar mass \( M_2O_3 \) is: \[ \text{Molar mass of } M_2O_3 = \frac{0.1595}{0.001} = 159.5 \text{ g/mol} \]

Step 5: Calculate Atomic Weight of the Metal

Use the formula for the molar mass of \( M_2O_3 \): \[2M + 3(16) = 159.5\] where 16 is the atomic weight of oxygen. \(3 \times 16 = 48\), so: \[2M + 48 = 159.5 \rightarrow 2M = 111.5 \rightarrow M = \frac{111.5}{2} = 55.75 \approx 55.8\]

Step 6: Find the Closest Option

The calculated atomic weight \(55.8\) matches with option (c).

Key Concepts

Reduction Reactions

In chemistry, a reduction reaction refers to the process where a substance gains electrons. This is usually alongside an oxidation reaction, where another substance loses electrons. Reduction is one half of a redox reaction with the other half being oxidation.
Reduction reactions often involve removing oxygen from a compound, or adding hydrogen to it, as in the case of burning hydrogen to form water.

• In our exercise, the metal oxide, \( M_2O_3 \), gets reduced to pure metal \( M \) and water \( H_2O \).
• The equation for this reduction reaction is: \( M_2O_3 + 3H_2 \rightarrow 2M + 3H_2O \).
• As hydrogen is added, the metal loses oxygen. This demonstrates the classical reduction by gaining hydrogen atoms.

Understanding reduction reactions is crucial for processes like smelting, where reduction is used to extract metals from their oxides.

Molar Mass Calculations

Calculating molar mass is a foundational skill in chemistry, which involves finding the mass of one mole of a substance. It is measured in grams per mole (g/mol). The molar mass of a compound provides the link between the mass of the substance and the amount in moles.

• In the given exercise, we first found the molar mass of \( H_2 \), which is approximately 2 g/mol.
• Next, calculating the molar mass of the metal oxide \( M_2O_3 \) was essential, as this gave us a deeper understanding of how much one mole of this compound would weigh.
• Using the formula: \[ \text{Molar mass of } M_2O_3 = \frac{\text{mass of } M_2O_3}{\text{moles of } M_2O_3} \]
• This led us to calculate a molar mass of 159.5 g/mol for \( M_2O_3 \).

This molar mass value is crucial for further calculations, particularly when determining the atomic weight of a single element in a compound.

Stoichiometry

Stoichiometry is the aspect of chemistry that looks at the quantitative relationships between reactants and products in a chemical reaction. It helps us understand how much of each reactant is needed to produce a desired amount of product.

• In this exercise, stoichiometry ensures the balanced equation \( M_2O_3 + 3H_2 \rightarrow 2M + 3H_2O \) is used to comprehend the mole ratios between reactants and products.
• The amount of hydrogen used was 0.003 moles, and according to stoichiometry, one mole of \( M_2O_3 \) reacts with three moles of \( H_2 \).
• This relationship allows us to determine the amount of \( M_2O_3 \) that reacted: \( \frac{0.003}{3} = 0.001 \) moles of \( M_2O_3 \).

Understanding stoichiometry helps in comprehending not just how elements and compounds react, but in predicting how much product will result from given reactants.

Chemical Formulas

Chemical formulas represent the elements present in a compound and the ratios in which they combine. They help in visualizing the structure and proportions of elements in a compound.

• For instance, the formula \( M_2O_3 \) tells us that two atoms of metal \( M \) are combined with three atoms of oxygen.
• These formulas are not only key for understanding what a compound consists of but also guide calculations. In this exercise, calculating the atomic weight of metal involved understanding the chemical formula \( M_2O_3 \).
• By using the sum of the atomic weights of all atoms in \( M_2O_3 \) (including those of the three oxygen atoms), we could isolate and compute the atomic weight of the metal \( M \).

Knowing how to interpret and use chemical formulas is vital for tasks like determining molecular weights, balancing reactions, and performing further calculations.