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

Since we know the mass and the number of molecules of X2, we can calculate the molecular mass and find out the identity of X. Molecular mass of X2 = (Mass of X2) / (Number of moles of X2) 1 mole = \(6.022 \times 10^{23}\) molecules. So, Number of moles of X2 = (Number of molecules of X2) / (Avogadro's number) = \(8.92 \times 10^{20}\) molecules / \(6.022 \times 10^{23}\) molecules/mole = 1.4818 x 10^(-3) moles of X2 Now, plug the values into the formula: Molecular mass of X2 = 0.105 g / 1.4818 x 10^(-3) moles = 70.8 g/mol Since there are two atoms of X in X2, the molar mass of X = 70.8 g/mol / 2 = 35.4 g/mol. The element with this approximate molar mass is Chlorine (Cl).

We know that in compound MX3, the mass percentage of X is 54.47%. Therefore, the mass percentage of M is 100% - 54.47% = 45.53%. Let the molar mass of one mole of MX3 be x g/mol. We can now write the equation using mass percentages and molar masses: (3 x (Molar mass of X) / (Molar mass of MX3)) x 100 = 54.47 (3 x 35.4) / x = 0.5447 Now, solve for x: x = (3 x 35.4) / 0.5447 x = 194.74 g/mol (Molar mass of MX3) To find the molar mass of M, subtract the mass of three moles of Cl: Molar mass of M = 194.74 g/mol - (3 x 35.4 g/mol) = 194.74 g/mol - 106.2 g/mol = 88.54 g/mol The element with this approximate molar mass is Strontium (Sr).

We have determined the identities of elements M and X as Strontium (Sr) and Chlorine (Cl), respectively. Therefore, the ionic compound MX3 can be written as SrCl3 and is named Strontium Chloride.

We start with 1g each of M (Strontium) and X2 (Chlorine), so we need to find the moles of M and X2: Moles of Strontium = mass / molar mass = 1/88.54 = 0.0113 moles Moles of Chlorine = mass / molar mass = 1/70.8 = 0.0141 moles Since the molar ratio of Sr:Cl in SrCl3 is 1:3, we need 0.0113 moles of Sr and 0.0339 moles of Cl. But we only have 0.0141 moles of Cl available. As Cl is present in a lesser amount, it is now the limiting reagent. Now, calculate the mass of SrCl3 that can be prepared using the limiting reagent Cl: moles of SrCl3 = moles of Cl / 3 mass of SrCl3 = moles of SrCl3 x molar mass of SrCl3 = (0.0141 / 3) x 194.74 = 0.919 g Thus, starting with 1g of Strontium and 1g of Chlorine, 0.919 g of Strontium Chloride (SrCl3) can be prepared.