Question

An element ‘A’ has a triiodide with the formula ${\mathbf{AI}}_{\mathbf{3}}$and a trichloride with the formula ${\mathbf{ACl}}_{\mathbf{3}}$. The triiodide is quantitatively converted to the trichloride when it is heated in a steam of chlorine, according to the reaction

role="math" localid="1663568551139" ${\mathbf{AI}}_{\mathbf{3}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{2}}{\mathbf{Cl}}_{\mathbf{2}}\to {\mathbf{AlCl}}_{\mathbf{3}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{2}}{\mathbf{I}}_{\mathbf{2}}$

If 0.8000g is treated, 0.3776g is obtained.

1. Calculate atomic mass of A.

1. Identify the element A.

Short answer

1. The atomic mass of A is 138.91g/mol.
2. The element A is probably lanthanum (La).

Step-by-step solution

Calculation of molar mass of AI3 and ACl3

Given equation is:

$A{I}_3+\frac{3}{2}C{l}_2\to AlC{l}_3+\frac{3}{2}{I}_2$

The mass of is 0.8000g and the mass of $AC{l}_3$is 0.3776g.

Let us assume the molar mass of ‘A’ to be ‘m’ g/mol.

Molar mass of$A{I}_3$ =$\left(m+3\times 126.90\right)g/mol$

Molar mass of$AC{l}_3=\left(m+3\times 35.45\right)g/mol$

Calculating the atomic mass of element ‘A’

As per the equation:

1 mole $A{I}_3$$\to$1 mole of $AC{l}_3$

Given: 0.800gdata-custom-editor="chemistry" $A{I}_3$$\to$ 0.3776g$AC{l}_3$

So, their moles (n) can be equated by the formula:

$\mathbf{n}\mathbf{=}\frac{\mathbf{weight}}{\mathbf{molarmass}}$

The known values are substituted.

$$\begin{gathered} \frac{0.800gA{I}_3}{\left(m+3\times 126.90\right)g/mol}=\frac{0.3776gAC{l}_3}{\left(m+3\times 35.45\right)g/mol} \\ 0.800\left(m+3\times 126.90\right)=0.3776\left(m+3\times 35.45\right) \\ 0.4224m=58.67 \\ m=138.9 \end{gathered}$$

Identification of element ‘A’

Lanthanum, La has the atomic mass of 138.91g/mol which is very close to mass of A (138.9 g/mol).

Therefore, the element A is probably Lanthanum (La).