Question

Determine the moles of AlCl3 produced when 4.75 grams of Cl₂ reacts with Al

according to the following reaction:
2
Al(s) + 3 Cl₂(g) →
2 AICI (S)

Answer 1

To determine the moles of AlCl3 produced when 4.75 grams of Cl₂ reacts with Al, divide the mass of Cl₂ by its molar mass to get the moles of Cl₂. Then, use the mole ratio from the balanced chemical equation to find the moles of AlCl₃ produced. We find that the moles of AlCl₃ produced when 4.75 grams of Cl₂ reacts with Al is 0.0475 moles.

Step-by-step explanation:

To determine the moles of AlCl3 produced when 4.75 grams of Cl₂ reacts with Al, we need to use the balanced chemical equation and the molar mass of Cl₂. The equation is: 2 Al(s) + 3 Cl₂(g) → 2 AlCl₃(s).

First, convert the mass of Cl₂ to moles by dividing the given mass by the molar mass of Cl₂ (70.90 g/mol).

This gives you the moles of Cl₂. Since the mole ratio of Cl₂ to AlCl₃ is 3:2, use that ratio to find the moles of AlCl₃ produced. Multiply the moles of Cl₂ by the ratio (2/3) to get the moles of AlCl₃.

Using the calculations, we find that the moles of AlCl₃ produced when 4.75 grams of Cl₂ reacts with Al is 0.0475 moles.