Question

N(AlCl3) = 185г. найти n(Al3+) = ? n(CL-) = ?

Answer 1

Там масса скорее всего
n(alcl3) =m/M=185/133.5=1.3857моль
Тогда n(al)=n(alcl3)
n(cl3) =n(al)*3=1.3857*3=4.157моль

Answer 2

To find the number of moles of aluminum ions (Al³⁺) and chloride ions (Cl⁻) in a given amount of aluminum chloride (AlCl₃), we need to use the molar mass and stoichiometry of the compound.

The molar mass of AlCl₃ can be calculated as follows:

Aluminum (Al) has a molar mass of 26.98 g/mol. Chlorine (Cl) has a molar mass of 35.45 g/mol (approximately).

The molar mass of AlCl₃ is:

Al: 1 atom × 26.98 g/mol = 26.98 g/mol Cl: 3 atoms × 35.45 g/mol = 106.35 g/mol

Total molar mass of AlCl₃ = 26.98 g/mol + 106.35 g/mol = 133.33 g/mol (approximately).

Now, we can calculate the number of moles using the formula:

Number of moles (n) = Mass (m) / Molar mass (M)

n(AlCl₃) = 185 g / 133.33 g/mol ≈ 1.388 mol

Since AlCl₃ dissociates into one aluminum ion (Al³⁺) and three chloride ions (Cl⁻) when it dissolves in water, the number of moles of Al³⁺ and Cl⁻ will be the same as the number of moles of AlCl₃.

Therefore: n(Al³⁺) ≈ 1.388 mol n(Cl⁻) ≈ 1.388 mol

So, the number of moles of aluminum ions (Al³⁺) and chloride ions (Cl⁻) is approximately 1.388 mol each.