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 \(\mathrm{X}_{2}\) contains \(8.92 \times 10^{20}\) molecules. The compound \(\mathrm{MX}_{3}\) consists of 54.47$\% \mathrm{X} \mathrm{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 \(\mathrm{MX}_{3} \mathrm{can}\) be prepared?

We know the mass of X2 (0.105 g) and are given the number of molecules in the sample (\(8.92\times10^{20}\)). To calculate the molar mass of X, we use Avogadro's number, which represents the number of entities (atoms or molecules) in one mole of a substance. Number of moles = \(\dfrac{Number\ of\ molecules}{Avogadro's\ number}\) $$ Moles\ of\ X2 = \frac{8.92\times10^{20}}{6.022\times10^{23}} \approx 1.482\times10^{-3}\ moles $$ Now, we will determine the molar mass of X2: Molar mass of X2 = \(\dfrac{Mass}{Moles}\) $$ Molar\ mass\ of\ X2 = \frac{0.105\ g}{1.482\times10^{-3}\ moles} \approx 70.8\ g/mol $$ Since X2 is diatomic, the molar mass of X will be half of the molar mass of X2: Molar mass of X = \(\dfrac{70.8\ g/mol}{2} = 35.4\ g/mol\) Using the periodic table, we see that X matches the element chlorine (Cl) with a molar mass of 35.4 g/mol.

We know that MX3 consists of 54.47% X by mass. Therefore, the mass percentage of M can be given as: Percent M = 100% - 54.47% = 45.53% Let's assume 100 g of MX3. In this case, the mass of M in MX3 will be 45.53 g and the mass of 3 Cl (3X) atoms will be 54.47 g. As we know the molar mass of Cl (35.4 g/mol), we can calculate the moles of Cl in 54.47 g: Moles of Cl = \(\dfrac{54.47\ g}{35.4\ g/mol} = 1.538\ moles\) Since there are three times the number of moles of Cl atoms as M atoms in MX3: Moles of M = \(\dfrac{1.538\ moles}{3} = 0.513\ moles\) Now, to find the molar mass of M, we can use its mass and moles: Molar mass of M = \(\dfrac{Mass}{Moles}\) $$ Molar\ mass\ of\ M = \frac{45.53\ g}{0.513\ moles} \approx 88.8\ g/mol $$ Using the periodic table, we see that M matches the element yttrium (Y) with a molar mass of 88.8 g/mol. So, the compound MX3 is Yttrium trichloride (YCl3).

We are given 1.00 g each of Y and Cl2. We will first calculate the moles of Y and Cl2: Moles of Y = \(\dfrac{1.00\ g}{88.8\ g/mol} = 0.0113\ moles\) Moles of Cl2 = \(\dfrac{1.00\ g}{70.8\ g/mol} = 0.0141\ moles\) Now, we will write the balanced chemical equation for the formation of YCl3: $$ 2Y + 3Cl2 \longrightarrow 2YCl3 $$ From the balanced equation, we can see that 3 moles of Cl2 (X2) are needed to react with 2 moles of Y to form 2 moles of YCl3. However, we do not have an equal ratio of moles as per the balanced equation. To identify the limiting reactant, we will divide the moles of Y and Cl2 by their respective stoichiometric coefficients: For Y: \(\dfrac{0.0113}{2} = 0.00565\) For Cl2: \(\dfrac{0.0141}{3} = 0.0047\) The smaller value (0.0047) corresponds to the limiting reactant, which in this case is Cl2. Now, we can use the moles of the limiting reactant to calculate the moles of YCl3 formed: Moles of YCl3 = \(0.0047 \times \dfrac{2\ moles\ YCl3}{3\ moles\ Cl2} = 0.00313\ moles\) Finally, we can calculate the mass of YCl3 formed using its molar mass (Y: 88.8 g/mol, 3Cl: 3 x 35.4 g/mol = 106.2 g/mol): Mass of YCl3 = \(0.00313\ moles \times (88.8\ g/mol + 106.2\ g/mol) \approx 0.609\ g\) So, the mass of YCl3 that can be prepared from 1.00 g each of Y and Cl2 is 0.609 g.