Question

An element A has a triiodide with the formula AI₃ and a trichloride with the formula ACl₃. The triiodide is quantitatively converted to the trichloride when it is heated in a stream of chlorine, according to the reaction.
AI₃ + 3/2 Cl₂ → ACl₃ + 3/2 I₂ If 0.8000 g AI₃ is treated, 0.3776 g ACl₃ is obtained.
Calculate the atomic mass of the element A. Identify the element A.

Answer 1

Answer:

The element A is Lanthanum, with a molar mass of 138.9g/mol

Explanation:

AI3 + 3/2 Cl2  →  3/2 I2 + ACl3

Compounds :

Al  ⇒ Molar mass = M

AI3 ⇒ Molar mass = M + 3*126.9

ACl3 ⇒Molar mass = M + 3*35.45

For 1 mole AI3 we have 1 mole ACl3

(0.800g * AI3) / ( M + 3*126,90)g/mole    =   (0.3776 g * ACl3) / (M + 3*35,45)g/mole

0.800 *(M + 3*35,45)  = 0.3776(M + 3*126,90)

0.8M  + 85.08 = 0.3776M + 143.75

0.8M - 0.3776M = 143.75-85.08

0.4224M = 58.67

M =138.9 g/mol

So A has a Molar mass of 138.9 g/mol ⇒ if we look at the periodic table we can find that Lanthanum (La) has a Molar mass of 138.9g/mol.