Question

A quantity of \(0.225 \mathrm{~g}\) of a metal \(\mathrm{M}\) (molar mass = 27.0 \(\mathrm{g} / \mathrm{mol}\) ) liberated \(0.303 \mathrm{~L}\) of molecular hydrogen (measured at \(17^{\circ} \mathrm{C}\) and \(741 \mathrm{mmHg}\) ) from an excess of hydrochloric acid. Deduce from these data the corresponding equation, and write formulas for the oxide and sulfate of \(\mathrm{M}\).

Short answer

The reaction equation is \( M + 2HCl \rightarrow MCl_2 + H_2 \); oxide: \( M_2O_3 \), sulfate: \( M_2(SO_4)_3 \).

Step-by-step solution

Calculate moles of metal M

We begin with calculating the moles of the metal \( M \) using its given mass and molar mass. Use the formula: \[\text{moles of } M = \frac{\text{mass}}{\text{molar mass}}\] Substitute the given values: \[\text{moles of } M = \frac{0.225 \, \text{g}}{27.0 \, \text{g/mol}} = 0.00833 \, \text{mol}\]

Correct gas volume to STP

The provided volume of hydrogen gas is at conditions \( 17^{\circ} \text{C} \) and \( 741 \text{mmHg} \). We use the ideal gas law to convert the gas volume to standard temperature and pressure (STP: 0°C, 1 atm). Use the combined gas law: \[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\] Convert initial conditions: \( V_1 = 0.303 \, \text{L}, \) \( P_1 = 741 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} \), \( T_1 = 273.15 + 17 = 290.15 \, \text{K} \).At STP: \( P_2 = 1 \, \text{atm}, \) \( T_2 = 273.15 \, \text{K} \). Solve for \( V_2 \):\[V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 0.303 \, \text{L} \times \frac{741}{760} \times \frac{273.15}{290.15} = 0.274 \, \text{L}\]

Calculate moles of H2 at STP

Using the corrected volume of \( H_2 \) at STP, calculate moles of \( H_2 \) using the molar volume of an ideal gas at STP (22.4 L/mol):\[\text{moles of } H_2 = \frac{0.274 \, \text{L}}{22.4 \, \text{L/mol}} = 0.0122 \, \text{mol}\]

Determine equivalence in reaction

Since metal \( M \) reacts with hydrochloric acid to produce hydrogen gas, we assume the reaction stoichiometry is \( M + 2HCl \rightarrow MCl_2 + H_2 \). This means 1 mole of \( M \) liberates 1 mole of \( H_2 \) gas.We have \( 0.0122 \, \text{mol} \) of \( H_2 \) produced, so \( 0.0122 \, \text{mol} \) of \( M \) should have been used as well. Since calculated moles of \( M \) used were 0.00833 (from Step 1), these should match, indicating stoichiometry equates \( M \) to \( H_2 \), and \( M \)'s likely oxidation state in \( MCl_2 \) is 2+

Deduce the oxide of M

An oxide with \( M \) in a 2+ oxidation state is likely \( M_2O_3 \) assuming \( M \) is in group 13, matching the equation \( 4M + 3O_2 \rightarrow 2M_2O_3 \).

Deduce the sulfate of M

Given \( M \) forms \( M_2O_3 \), and presuming \( M \) reacts typically to form sulfates in oxidation 2+ state, the sulfate would be \( M_2(SO_4)_3 \).

Key concepts

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction. In this exercise, stoichiometry helps us understand how much of the metal M reacts with hydrochloric acid to produce a specific amount of hydrogen gas. By analyzing the balanced equation, which is often seen as \(M + 2HCl \rightarrow MCl_2 + H_2\), we determine that one mole of metal M yields one mole of hydrogen gas.
This stoichiometric ratio is vital for calculating how many moles of reactants are consumed and how many moles of products are formed in reactions. It serves as a groundwork to link the "recipe" of the chemical reaction to actual amounts needed or produced, measured typically in grams or liters. Understanding stoichiometry is essential for any chemical calculations, especially when aiming to achieve precise and expected results in laboratories.

Ideal Gas Law

The ideal gas law is an important equation used to understand the behavior of gases under various conditions. The law is represented as \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.
In this exercise, the ideal gas law is used to adjust the volume of hydrogen gas produced to standard temperature and pressure (STP) conditions, making subsequent calculations straightforward. Using adjusted conditions simplifies calculations and allows us to accurately find the number of moles of hydrogen, assisting in linking it to moles of metal M used through stoichiometry.

• Convert given conditions to STP to standardize calculations.
• The temperature must be in Kelvin when using the ideal gas law.
• The ideal gas law helps relate measurable properties of a gas to its moles.

Grasping this law is crucial for accurately determining aspects like gas volume and conditions in both experimental and theoretical chemistry.

Molar Mass

Molar mass is the mass of one mole of a substance, typically measured in grams per mole (g/mol). This concept is crucial because it allows conversion between the mass of a substance and the number of moles, bridging the macroscopic and molecular worlds.
In this problem, the molar mass of metal M is given as 27.0 g/mol. By dividing the mass of the metal by its molar mass, we find the number of moles. This number helps determine quantities when dealing with chemical reactions, as seen when calculating moles of reactants or products.

• Molar mass is derived from atomic masses on the periodic table.
• It's used as a conversion factor in stoichiometric calculations.
• Understanding molar mass is crucial for laboratory measurements and scaling reactions.

Knowing molar mass allows chemists to predict how much substance is needed to react with another or to produce a given amount of product.

Oxidation States

Oxidation states, also known as oxidation numbers, help describe the transfer of electrons in a chemical reaction. They indicate the degree of oxidation (loss of electrons) for an atom in a compound. In simple terms, it shows how much electron control an atom has in a bond.
For metal M reacting with hydrochloric acid, it's assumed that the oxidation state changes to 2+ based on the products formed, such as \(MCl_2\). This 2+ state suggests that each atom of metal M loses two electrons.

• Oxidation states can determine how atoms will interact in a reaction.
• They are crucial for balancing redox reactions.
• Knowing these states helps predict product formation and stability.

Understanding oxidation states is vital for writing chemical equations correctly and predicting how substances will react with each other.