Question

A metal oxide has the formula \(X_{2} \mathrm{O}_{3}\). It can be reduced by hydrogen to give free metal and water. \(0.1596 \mathrm{~g}\) of metal oxide requires \(6 \mathrm{mg}\) of hydrogen for complete reduction. The atomic weight of the metal (in amu) is: (a) \(15.58\) (b) \(155.8\) (c) \(5.58\) (d) \(55.8\)

Short answer

The atomic weight of the metal (X) is approximately 20.095 amu, which is not listed among the given options.

Step-by-step solution

Write the chemical reaction

Write the balanced chemical equation for the reduction of the metal oxide by hydrogen. Given that the metal oxide is represented by \(X_2O_3\) and it reacts with hydrogen (H2) to give the metal (X) and water (H2O). The balanced chemical equation is:\[X_2O_3 + 3H_2 \rightarrow 2X + 3H_2O\]

Convert hydrogen mass to moles

Convert the mass of hydrogen required for the reaction from milligrams to grams and then to moles using the molar mass of hydrogen (which is approximately 1.008 grams per mole for H2).\[6 \text{ mg} = 0.006 \text{ g}\]\[\text{Moles of } H_2 = \frac{0.006 \text{ g}}{2 \times 1.008 \text{ g/mol}} \approx 0.00297619 \text{ moles}\]

Determine the moles of metal

Use the stoichiometry from the balanced chemical equation to find the moles of metal produced. The mole ratio of hydrogen to metal is 3:2. Therefore, for every 3 moles of hydrogen, 2 moles of metal are produced:\[\text{Moles of } X = \frac{2}{3} \times \text{Moles of } H_2 \approx \frac{2}{3} \times 0.00297619 \approx 0.00198413 \text{ moles}\]

Calculate the molar mass of metal

Calculate the molar mass of the metal using the mass of the metal oxide and the moles of metal obtained.\[\text{Molar mass of } X = \frac{\text{Mass of } X_2O_3}{2 \times \text{Moles of } X} \]\[\text{Molar mass of } X = \frac{0.1596 \text{ g}}{2 \times 0.00198413 \text{ moles}} \approx 40.19 \text{ g/mol}\]

Determine the atomic weight of the metal

Since the molar mass found is for \(X_2\), the atomic weight of metal \(X\) will be half of that value.\[\text{Atomic weight of } X = \frac{40.19 \text{ g/mol}}{2} \approx 20.095 \text{ g/mol or amu}\]None of the given options match this result, suggesting either a mistake in the calculation or an error in the provided options.

Key concepts

Chemical Reaction Balancing

Understanding chemical reaction balancing is fundamental in solving stoichiometry problems, including those you might encounter on the JEE. Let's take the provided exercise as an example. The reaction in question was the reduction of a metal oxide, denoted as \( X_2O_3 \), by hydrogen gas (\( H_2 \)). The key to balancing a chemical reaction is ensuring that the number of atoms for each element is the same on both the reactant and product sides of the equation.

In this case, the balanced equation is \( X_2O_3 + 3H_2 \rightarrow 2X + 3H_2O \). Notice how the oxygen atoms are balanced (three on each side) as are the atoms of the metal, \( X \), and hydrogen. For learners, a useful tip is to start balancing elements that appear in the least number of compounds and leave hydrogen and oxygen for last, as they are generally more prevalent throughout the reaction.

Molar Mass Calculation

Molar mass calculation is a pivotal skill in chemistry, especially when working through stoichiometry problems. It represents the mass of one mole of a substance (usually in grams per mole). In the context of JEE problems, such as the one provided, once you've balanced the chemical equation, you can use the stoichiometry to find the moles of reactants and products.

For example, you're given the mass of hydrogen gas and want to find out the molar mass of the metal in the oxide \(X_2O_3\). By converting the mass of hydrogen to moles (using its molar mass of approximately 2.016 grams per mole for \(H_2\)), you can then apply the stoichiometry of the balanced equation to find the moles of metal. With the moles of metal and the given mass of metal oxide, you calculate the molar mass. Accurate molar mass calculations are crucial as they impact subsequent steps, like determining the atomic weight of the metal.

Reduction Reactions

Reduction reactions are chemical processes where a substance gains electrons, typically resulting in a decrease in its oxidation state. In stoichiometry and JEE exams, understanding these reactions is necessary for predicting products and reactants. In our exercise, hydrogen gas reduces the metal oxide to free metal, illustrating a classic reduction.

During this type of reaction, it's important to note the transfer of electrons and changes in oxidation states. For example, in our reaction, the metal in the oxide is being reduced as it goes from being part of an ionic compound to its elemental state. It's helpful for students to remember that reduction usually occurs with a reductant, in this case, hydrogen, which gets oxidized in the process, as seen in the production of water. This interplay is at the heart of redox reactions and is essential for correctly solving problems involving stoichiometry and chemical changes.