The compound As \(_{2} \mathrm{L}_{4}\) is synthesized by reaction of arsenic metal with arsenic triodide. If a solid cubic block of arsenic $\left(d=5.72 \mathrm{g} / \mathrm{cm}^{3}\right)\( that is 3.00 \)\mathrm{cm}$ on edge is allowed to react with \(1.01 \times 10^{24}\) molecules of arsenic triodide, what mass of \(\mathrm{As}_{2} \mathrm{L}_{4}\) can be prepared? If the percent yield of \(\mathrm{As}_{2} \mathrm{L}_{4}\) was 75.6\(\%\) what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was actually isolated?

Since the density, d = mass/volume, we can calculate the mass of the solid arsenic cube by using the given dimensions: mass = density × volume mass = 5.72 g/cm³ × (3.00 cm)³ mass = 5.72 g/cm³ × 27.00 cm³ mass ≈ 154.44 g The mass of the solid arsenic cube is 154.44 g.

To calculate the moles of arsenic from the mass, we'll use the formula: moles = mass/molar_mass, where the molar mass of As = 74.92 g/mol. moles of As = 154.44 g / 74.92 g/mol ≈ 2.06 mol We are given the number of molecules of arsenic triiodide, 1.01 × 10²⁴. To convert this to moles, we'll use Avogadro's number (6.022 × 10²³ molecules/mol): moles of AsI₃ = (1.01 × 10²⁴) / (6.022 × 10²³) moles of AsI₃ ≈ 1.68 mol

First, we must determine the stoichiometry between As and AsI₃ based on the chemical reaction. Since 1 mole of As reacts with 1 mole of AsI₃ to produce 1 mole of As\(_{2}\)I\(_{4}\), we have: As + AsI₃ → As\(_{2}\)I\(_{4}\) Now since As is in excess, we will base our calculations on the moles of AsI₃: moles of AsI₃ = 1.68 mol Now, we can find the theoretical mass of As\(_{2}\)I\(_{4}\) produced, using its molar mass (2 moles of As = 2 × 74.92 g/mol, and 4 moles of I = 4 × 126.90 g/mol): molar_mass of As\(_{2}\)I\(_{4}\) = 2 × 74.92 g/mol + 4 × 126.90 g/mol = 656.84 g/mol mass of As\(_{2}\)I\(_{4}\) = moles of As\(_{2}\)I\(_{4}\) × molar_mass of As\(_{2}\)I\(_{4}\) mass of As\(_{2}\)I\(_{4}\) = 1.68 mol × 656.84 g/mol ≈ 1103.47 g The theoretical mass of As\(_{2}\)I\(_{4}\) that can be prepared is approximately 1103.47 g.

The percent yield of As\(_{2}\)I\(_{4}\) is given as 75.6%. We can use this and the theoretical mass to calculate the actual mass of As\(_{2}\)I\(_{4}\) isolated: actual_mass = (percent_yield/100) × theoretical_mass actual_mass = 0.756 × 1103.47 g actual_mass ≈ 833.62 g The actual mass of As\(_{2}\)I\(_{4}\) isolated is approximately 833.62 g.