An ionic compound \(\mathrm{MX}_{3}\) is prepared according to the following unbalanced chemical equation. $$\mathbf{M}+\mathbf{X}_{2} \longrightarrow \mathbf{M X}_{3}$$ A \(0.105-\mathrm{g}\) sample of \(\mathrm{X}_{2}\) contains \(8.92 \times 10^{20}\) molecules. The compound \(\mathrm{MX}_{3}\) consists of \(54.47 \%\) X by mass. What are the identities of \(\mathrm{M}\) and \(\mathrm{X}\), and what is the correct name for \(\mathrm{MX}_{3} ?\) Starting with 1.00 g each of \(M\) and \(X_{2}\), what mass of \(M X_{3}\) can be prepared?

To calculate the molar mass of X₂, we'll use the mass of X₂ and the number of molecules available: Molar mass of X₂ = (Mass of X₂) / (Number of molecules of X₂ / Avogadro's number) Avogadro's number = 6.022 × 10²³ Number of moles of X₂ = (8.92 × 10²⁰ molecules) / (6.022 × 10²³) = 0.001482 moles Molar mass of X₂ = 0.105 g / 0.001482 moles = 70.81 g/mol

Since X₂ is a diatomic molecule, the atomic mass of X can be calculated by dividing the molar mass of X₂ by 2: Atomic mass of X = Molar mass of X₂ / 2 Atomic mass of X = 70.81 g/mol / 2 = 35.40 g/mol

We know that the mass of X in MX₃ is 54.47%. Therefore, the mass of M in MX₃ will be 100% - 54.47% = 45.53%. Let molar mass of MX₃ = A g/mol Mass of X in 1 mol of MX₃ = 3 × atomic mass of X = 3 × 35.40 g/mol = 106.20 g/mol Mass of M in 1 mol MX₃ = mass of M in MX₃ = A × 45.53% Now, mass of X in MX₃ = A × 54.47% 106.20 g/mol = A × 54.47% We can solve for A: A = 106.20 g/mol / 54.47% = 194.95 g/mol Now that we have the molar mass of MX₃, we can determine the mass of M in 1 mol of MX₃ and identify M and X: Mass of M = A × 45.53% = 194.95 g/mol × 45.53% = 88.75 g/mol We can now identify M and X: M ≈ 89 g/mol (Rb, Rubidium) X ≈ 35.5 g/mol (Cl, Chlorine) Thus, we have found the elements in MX₃ to be Rubidium and Chlorine, so the compound is Rubidium Chloride (RbCl₃).

We start with 1.00 g of M (Rb) and 1.00 g of X₂ (Cl₂). Since M and X₂ react in a 1:1 ratio, we can find which reactant is the limiting reactant. Number of moles of M = 1.00 g / (atomic mass of Rb) = 1.00 g / 88.75 g/mol = 0.01127 moles Number of moles of X₂ = 1.00 g / (molar mass of Cl₂) = 1.00 g / 70.81 g/mol = 0.01412 moles Since the ratio of M:X₂ in the reaction is 1:1, Rb is the limiting reactant. Now, for each mole of Rb, one mole of RbCl₃ is formed. So, 0.01127 moles of Rb reacts completely forming 0.01127 moles of RbCl３. Let's calculate the mass of RbCl₃ formed: Mass of RbCl₃ = 0.01127 moles × (molar mass of RbCl₃) = 0.01127 moles × 194.95 g/mol = 2.20 g Hence, we can prepare 2.20 g of RbCl₃ using 1.00 g each of Rb and Cl₂.