Question

Consider the balanced chemical equation when 18.3 g Al is reacted with 113 g I₂ to form AlI₃(g). Calculate the theoretical yield in from the complete reaction of according to the following balanced chemical equation: Al(s) + I₂(s) → AlI₃(s)

Answer 1

When 17.467 grams of chlorine gas reacts with excess aluminum, approximately 21.87 grams of solid aluminum chloride will be formed.

Step-by-step explanation:

To determine the mass of solid aluminum chloride formed when 17.467 grams of chlorine gas reacts with excess aluminum, we need to convert the given mass of chlorine gas to moles and then use the stoichiometric ratio from the balanced chemical equation.

The molar mass of Cl2 is 70.90 g/mol, so we convert the mass of chlorine gas to moles by dividing 17.467 g by 70.90 g/mol, giving us approximately 0.246 moles of Cl2.

From the balanced chemical equation 2 Al (s) + 3 Cl2 (g) → 2AlCl3 (s), we can see that for every 3 moles of Cl2, 2 moles of AlCl3 are formed. Therefore, we can calculate the moles of AlCl3 produced by multiplying the moles of Cl2 by the stoichiometric ratio, resulting in 0.246 moles * (2 moles AlCl3 / 3 moles Cl2) = 0.164 moles of AlCl3.

To find the mass of solid aluminum chloride formed, we multiply the number of moles of AlCl3 by its molar mass, which is 133.33 g/mol. Therefore, the mass of solid aluminum chloride formed is approximately 0.164 moles * 133.33 g/mol = 21.87 grams.