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. \(27.9\) b. \(159.6\) c. \(79.8\) d. \(55.8\)

Short answer

The atomic weight of the metal is 55.8.

Step-by-step solution

Write the balanced chemical equation

The metal oxide has the formula \(Z_2O_3\). The balanced chemical equation for its reduction by hydrogen is: \(Z_2O_3 + 3H_2 \rightarrow 2Z + 3H_2O\).

Calculate moles of hydrogen used

Given the mass of hydrogen is \(6\text{mg} = 0.006\text{g}\). The molar mass of hydrogen is \(2\text{g/mol}\). Thus, the moles of hydrogen used is \(\frac{0.006}{2} = 0.003\text{ moles}\).

Use stoichiometry to calculate moles of metal oxide

From the equation, \(3\) moles of \(H_2\) reduce \(1\) mole of \(Z_2O_3\). Thus, \(0.003\) moles of \(H_2\) would reduce \(\frac{0.003}{3} = 0.001\) moles of \(Z_2O_3\).

Calculate molar mass of the metal oxide

The mass of \(Z_2O_3\) given is \(0.1596\text{g}\). Thus the molar mass is \(\frac{0.1596}{0.001} = 159.6\text{ g/mol}\).

Determine the molar mass of metal (Z)

Since the formula of the metal oxide is \(Z_2O_3\), its molar mass can also be expressed as \(2M_Z + 3(16)\). Setting this equal to the molar mass of the oxide calculated, we have \(2M_Z + 48 = 159.6\). Solving for \(M_Z\) gives \(M_Z = \frac{159.6 - 48}{2} = 55.8\text{ g/mol}\).

Key concepts

Chemical Equations

In chemistry, reactions are typically described using chemical equations. These equations provide a way to represent the transformation of reactants into products. For the reduction process of metal oxides, like the one given in the problem, equations play a crucial role in visualizing the chemical process involved. For example, the reduction of the metal oxide \( Z_2O_3 \) by hydrogen is shown in the balanced equation: \(Z_2O_3 + 3H_2 \rightarrow 2Z + 3H_2O\). This equation demonstrates how one mole of \( Z_2O_3 \) reacts with three moles of hydrogen to produce two moles of the free metal, \( Z \), and three moles of water.

• The symbols and numbers give a precise understanding of which elements participate in the reaction.
• The coefficients before each molecule balance the equation, ensuring that the number of atoms for each element is the same on both sides.
• This balanced equation is crucial for performing calculations related to stoichiometry.

Understanding how to read and balance chemical equations is a key skill when studying stoichiometry, as it enables the correct interpretation of chemical reactions.

Molar Mass Calculation

Calculating molar mass is an essential step when solving stoichiometry problems. Molar mass, expressed in g/mol, tells us the mass of one mole of a substance, which we need for converting between mass and moles.For example, in this exercise, you are given the mass of hydrogen as 6 mg. First, convert it to grams, which is 0.006 g. Given hydrogen's molar mass is \(2 \text{ g/mol}\), you can use the formula: \[\text{Moles of } H_2 = \frac{\text{mass of } H_2}{\text{molar mass of } H_2} = \frac{0.006}{2} = 0.003 \text{ moles}\]Once you have the moles of hydrogen, you use the stoichiometric relationship from the chemical equation to find the moles of the metal oxide \( Z_2O_3 \). Ultimately, the goal is to determine the molar mass of the metal \( Z \). After calculating the molar mass of the metal oxide as \( 159.6 \text{ g/mol} \) through conversions, use this and the molecular formula \( Z_2O_3 \) (where the oxygen's contribution is known) to determine the metal's molar mass:\[2M_Z + 48 = 159.6\]Solving gives \( M_Z = 55.8 \text{ g/mol} \).

• Use known molar masses of simple elements to find complex compounds.
• These calculations link measured mass to chemical formulas, providing deeper insights.

Reduction Reactions

Reduction reactions are a type of chemical reaction that involves the gain of electrons by an atom or molecule. In stoichiometry, reduction often goes hand-in-hand with oxidation reactions, hence the name redox reactions. However, in this exercise, you are mainly focused on the reduction process. In the context of the problem described, the metal oxide \( Z_2O_3 \) being reduced means that it gains electrons, allowing the metal component \( Z \) to be freed from the oxide complex. This is facilitated by hydrogen, which acts as the reducing agent, donating electrons and transforming into water:\[ Z_2O_3 + 3H_2 \rightarrow 2Z + 3H_2O \]

• The reduction of the metal oxide to pure metal illustrates how common metals are often found in nature as oxides.
• Reduction reactions are key in metallurgy, involving transforming mined oxides to usable metal.

Understanding reduction reactions is essential for the real-world application of many industrial processes, including the production of pure metals from their ores.