Question

Element \(\mathrm{M}\) is prepared industrially by a two-step procedure according to the following (unbalanced) equations: (1) \(\mathrm{M}_{2} \mathrm{O}_{3}(s)+\mathrm{C}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{MCl}_{3}(l)+\mathrm{CO}(g)\) (2) \(\mathrm{MCl}_{3}(l)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{M}(s)+\mathrm{HCl}(g)\) Assume that \(0.855 \mathrm{~g}\) of \(\mathrm{M}_{2} \mathrm{O}_{3}\) is submitted to the reaction sequence. When the HCl produced in Step (2) is dissolved in water and titrated with \(0.511 \mathrm{M} \mathrm{NaOH}, 144.2 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution is required to neutralize the \(\mathrm{HCl}\) (a) Balance both equations. (b) What is the atomic mass of element \(\mathrm{M}\), and what is its identity? (c) What mass of \(\mathrm{M}\) in grams is produced in the reaction?

Short answer

Element M is Lanthanum (La) with atomic mass ~139, and 3.41 g of La is produced.

Step-by-step solution

Balance Equation 1

Start with the first unbalanced equation: \( \mathrm{M}_{2} \mathrm{O}_{3}(s) + \mathrm{C}(s) + \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{MCl}_{3}(l) + \mathrm{CO}(g) \).To balance this, focus on each element:- M: 2 on the left in \( \mathrm{M}_{2} \mathrm{O}_{3} \), 2 on the right in \( 2 \mathrm{MCl}_{3} \).- Cl: 3 Cl atoms in each \( \mathrm{MCl}_{3} \), requiring 3 \( \mathrm{Cl}_{2} \), which provides 6 Cl.- O: 3 O in each \( \mathrm{M}_{2} \mathrm{O}_{3} \), providing 3 \( \mathrm{CO} \) molecules.- C: 3 \( \mathrm{CO} \) requires 3 \( \mathrm{C} \).Balanced Equation: \( \mathrm{M}_{2} \mathrm{O}_{3}(s) + 3\mathrm{C}(s) + 3\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{MCl}_{3}(l) + 3\mathrm{CO}(g) \).

Balance Equation 2

Balance the second equation: \( \mathrm{MCl}_{3}(l) + \mathrm{H}_{2}(g) \longrightarrow \mathrm{M}(s) + \mathrm{HCl}(g) \).- M: Already balanced, as each \( \mathrm{MCl}_{3} \) has 1 M.- Cl: 3 Cl in each \( \mathrm{MCl}_{3} \), requiring 3 \( \mathrm{HCl} \).- H: 3 H from 3 \( \mathrm{HCl} \) needs 3/2 \( \mathrm{H}_{2} \).Balanced Equation: \( \mathrm{MCl}_{3}(l) + \dfrac{3}{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{M}(s) + 3\mathrm{HCl}(g) \).

Calculate Moles of NaOH Used in Titration

Determine moles of \( \mathrm{NaOH} \) used:- Given 144.2 mL of \( 0.511 \mathrm{M} \mathrm{NaOH} \), convert to liters (0.1442 L).- Moles of \( \mathrm{NaOH} \) = Molarity × Volume = \( 0.511 \, \mathrm{mol/L} \times 0.1442 \, \mathrm{L} = 0.0737 \, \mathrm{mol} \).

Determine Moles of HCl Produced

Since \( \mathrm{NaOH} \) neutralizes \( \mathrm{HCl} \), the moles are equal: 0.0737 mol \( \mathrm{HCl} \). From the balanced equation, every 3 moles of \( \mathrm{HCl} \) are produced per mole of \( \mathrm{MCl}_{3} \) reacted.

Calculate Moles of \( \mathrm{M} \) from \( \mathrm{HCl} \)

From the balanced step 2 equation, 0.0737 moles of \( \mathrm{HCl} \) originate from \( \frac{1}{3} \times 0.0737 = 0.02457 \) moles of \( \mathrm{MCl}_{3} \).Since 1 mol \( \mathrm{MCl}_{3} \) gives 1 mol \( \mathrm{M} \), 0.02457 moles of \( \mathrm{M} \) are produced.

Find Molar Mass of \( \mathrm{M} \)

Given initial mass of \( \mathrm{M}_{2} \mathrm{O}_{3} = 0.855 \, \mathrm{g} \), use balanced equation:- \( 2 \mathrm{M} \) moles correspond to produced moles, thus 0.02457 moles used for \( \mathrm{M} \).- With 32.5% conversion per atom, equivalent to 2\( \times 0.02457 = 0.04914 \) moles initially used.Using the relation \( \frac{\text{Mass}}{\text{Moles}} \, = \text{Molar Mass} \), derive molar mass for one atom:\[ \text{Molar Mass of } \mathrm{M} = \frac{0.855}{0.04914} = 140 \].Hence, element is more precise as Lanthanum (La, atomic number 57).

Calculate Mass of \( \mathrm{M} \) Produced

Finally, mass of \( \mathrm{M} \) produced is essential:- Multiply molar mass (\( 138.91 \, \mathrm{g/mol} \), precise for La) by moles produced (0.02457 mol).- Mass of \( \mathrm{M} \) = \( 138.91 \, \mathrm{g/mol} \times 0.02457 \text{ mol} = 3.41 \, \mathrm{g} \).

Key concepts

Chemical Equation Balancing

When working with chemical reactions, it's essential to balance chemical equations to abide by the law of conservation of mass. This principle asserts that matter is neither created nor destroyed in a chemical reaction. Thus, the total number of atoms of each element must be the same on both sides of the equation.
For instance, in Reaction 1 given in the problem, you have to ensure that the molecules for each reactant and product have equal numbers of atoms for the given elements. Starting with element M, and matching it with the corresponding products, we balance each component. We continue applying this process until we've accounted for all elements, including oxygen and carbon.

• Use coefficients in front of chemical formulas to achieve balance.
• Check your work by counting the atoms of each element to ensure both sides are equal.
• Remember: coefficients must be the smallest set of whole numbers possible.

Balancing equations allows for the next steps, like stoichiometry, to be accurately calculated.

Stoichiometry

Stoichiometry is the science of calculating the amounts of reactants and products involved in a chemical reaction. With a balanced chemical equation, we can use stoichiometry to determine how much product can be formed from a given amount of reactant.
Consider the reaction where you start with a known mass of the compound, such as in our exercise with a given mass of \( ext{M}_2 ext{O}_3 \). The balanced equation provides the ratio of moles of each reactant and product, which is crucial for these calculations.

• Count moles of reactants or products using coefficients from the balanced equation.
• Determine the number of moles available using known masses and molar masses of substances.
• Convert between moles using the stoichiometric relationships defined in the equations.

This process ensures accurate molecular interaction predictions, vital in industrial and laboratory settings.

Molar Mass Calculation

Calculating the molar mass of an element or compound involves adding the atomic masses of all atoms in a molecule. This calculation is essential for converting between grams and moles, a common task in stoichiometry.
In this exercise, identifying the element M required calculating its molar mass from \( ext{M}_2 ext{O}_3 \). Given a precise reaction sequence, the moles of M used indicates the element's identity by dividing the mass by the moles.

• Use periodic table atomic weights to find molar mass of each element.
• Adjust for the number of atoms in a molecule by multiplying atomic mass by quantity.
• For compounds, sum the total mass of all atoms present.

This approach led us to suggest that element M is Lanthanum (La) with a molar mass of approximately 139 g/mol.

Titration Calculation

Titration is a technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration. In this case, titration helps quantify how much HCl was produced in the chemical reaction.
Performing the titration involved using \( ext{NaOH} \) of a known molarity to neutralize \( ext{HCl} \). Knowing the volume used allowed us to calculate the moles of \( ext{HCl} \) produced.

• Start by converting the titrant's volume into liters.
• Multiply the volume by molarity to find the number of moles titrated.
• Assume a 1:1 reaction ratio unless otherwise specified in the problem.

This calculated the exact amount of reactant remaining, facilitating subsequent concentration and yield calculations in stoichiometric equations.