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) \(58.5\)

Short answer

The atomic weight of the metal is closest to 55.8.

Step-by-step solution

Write the Reaction Equation

The reduction of the metal oxide can be described by the chemical equation: \[ \text{M}_2\text{O}_3 + 3\text{H}_2 \rightarrow 2\text{M} + 3\text{H}_2\text{O} \] This shows that 3 moles of hydrogen react with 1 mole of metal oxide to produce 2 moles of metal.

Calculate Moles of Hydrogen

Given mass of hydrogen used is 6 mg or 0.006 g. Using the molar mass of hydrogen (2 g/mol), calculate the moles of hydrogen: \[ \text{Moles of } \text{H}_2 = \frac{0.006}{2} = 0.003 \text{ moles} \]

Determine Moles of Metal Oxide Reduced

From the balanced chemical equation, 3 moles of \(\text{H}_2\) reduces 1 mole of \(\text{M}_2\text{O}_3\). So, \[ \text{Moles of } \text{M}_2\text{O}_3 = \frac{0.003}{3} = 0.001 \text{ moles} \]

Determine Molar Mass of Metal Oxide

The mass of the metal oxide given is 0.1595 g. Using the number of moles calculated, the molar mass of \(\text{M}_2\text{O}_3\) is: \[ \text{Molar mass of } \text{M}_2\text{O}_3 = \frac{0.1595}{0.001} = 159.5 \text{ g/mol} \]

Calculate Atomic Weight of Metal

Assume the atomic weight of metal is \(M\). Then the formula weight of \(\text{M}_2\text{O}_3\) is \(2M + 3(16)\). Set this equal to the molar mass: \[ 2M + 48 = 159.5 \] \[ 2M = 111.5 \] \[ M = \frac{111.5}{2} = 55.75 \]

Choose the Nearest Option

The calculated atomic weight of the metal is 55.75. Among the given options, this is closest to (c) 55.8.

Key concepts

Metal Oxide Reduction

Metal oxide reduction involves removing oxygen from a metal compound to release the pure metal. In many chemical processes, hydrogen is used as the reducing agent since it effectively attaches to the oxygen, forming water. In the case of this exercise, the metal oxide with the formula \( \mathrm{M}_2\mathrm{O}_3 \) reacts with hydrogen to produce the free metal and water.

• In the reaction: \( \mathrm{M}_2\mathrm{O}_3 + 3\mathrm{H}_2 \rightarrow 2\mathrm{M} + 3\mathrm{H}_2\mathrm{O} \), hydrogen takes the oxygen away from the metal oxide.
• The balanced equation shows how every mole of metal oxide requires 3 moles of hydrogen to fully reduce and release 2 moles of metal.
• The water formation indicates complete reduction.

This simple chemical transformation is key in metallurgy for turning ores into refined metals.

Stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions. By understanding the ratios in which chemicals react, we can determine the amounts of reactants required or products formed. In this problem, stoichiometry helps us assess how much hydrogen is needed to react with the metal oxide.

• The equation \( \mathrm{M}_2\mathrm{O}_3 + 3\mathrm{H}_2 \rightarrow 2\mathrm{M} + 3\mathrm{H}_2\mathrm{O} \) gives us the mole ratios needed for these calculations.
• From the balanced equation, we see that 1 mole of \( \mathrm{M}_2\mathrm{O}_3 \) requires 3 moles of \( \mathrm{H}_2 \) for complete conversion.
• These stoichiometric coefficients dictate how much of one substance reacts with or is produced from another.

Stoichiometry is essential for predicting the outcomes of reactions accurately.

Chemical Equation Balancing

Balancing chemical equations is crucial for accurately representing chemical reactions. It ensures the same number of each atom is present on both sides of the equation. Balancing the metal oxide reduction equation allows us to determine precisely how reactants convert to products.

• The equation: \( \mathrm{M}_2\mathrm{O}_3 + 3\mathrm{H}_2 \rightarrow 2\mathrm{M} + 3\mathrm{H}_2\mathrm{O} \) shows that each element is balanced.
• There are 2 metal atoms, 3 oxygen atoms, and 6 hydrogen atoms on each side of the equation.
• This balance is critical to correctly apply stoichiometric principles to solve quantitative chemical problems.

Balancing assures that mass and atoms remain conserved throughout the chemical reaction.

Molar Mass Calculation

Calculating molar mass involves determining the mass of one mole of a substance. It depends on the combined atomic masses of the atoms within a compound. Accurate molar mass calculation allows for the determination of quantities in reactions.

• In this exercise, we calculated the molar mass of \( \mathrm{M}_2\mathrm{O}_3 \) using its mass and moles: \( 159.5 \text{ g/mol} \).
• This calculation requires knowing the atomic weights of constituent elements and combining them, as seen in forming the relation: \( 2M + 48 \).
• Once the molar mass is known, it allows further calculation, like finding the atomic weight of the metal involved.

Understanding molar mass helps translate between moles, molecules, and grams, a fundamental aspect of chemistry problem-solving.