\[ \text{Ga}_2\text{O}_3(s) + 3\text{SOCl}_2(l) \rightarrow 2\text{GaCl}_3(s) + 3\text{SO}_2(g) \]

In a certain reaction, 35.9 g of \(\text{Ga}_2\text{O}_3\) reacts with 55.4 g of \(\text{SOCl}_2\). The \(\text{GaCl}_3\) produced is collected, and its mass is found to be 48.83 g.

What is the theoretical yield of \(\text{GaCl}_3\)?

The theoretical yield of GaCl3, calculated using stoichiometry and the balanced equation, is 67.51 g. This represents the maximum amount that could be produced from the reactants provided.

Explanation:

The question asks for the theoretical yield of GaCl3 when Ga2O3 reacts with SOCl2. To find the theoretical yield, we'll use stoichiometry based on the balanced chemical equation. First, we calculate the moles of Ga2O3 using its molar mass (187.44 g/mol). Then, we use the molar ratio from the balanced equation to find the moles of GaCl3. Finally, we convert the moles of GaCl3 to grams using its molar mass (176.32 g/mol). Here's the calculation:

35.9 g Ga2O3 x (1 mol Ga2O3 / 187.44 g) = 0.1915 mol Ga2O3
0.1915 mol Ga2O3 x (2 mol GaCl3 / 1 mol Ga2O3) = 0.383 mol GaCl3
0.383 mol GaCl3 x (176.32 g GaCl3 / 1 mol) = 67.51 g GaCl3

The theoretical yield of GaCl3 is therefore 67.51 g. This is the maximum amount of GaCl3 that can be produced from the given amounts of reactants if the reaction goes to completion without any losses.