Question

Identify the right option:
In an Experiment 6.67 g of $\operatorname{Al}{\text{C}}{l_3}$ was produced and 0.54 g of Al remained untreated. How many grams of Al and ${\text{C}}{{\text{l}}_{\text{2}}}$ were taken originally.
A) 0.07, 0.15
B) 0.07, 0.05
C) 0.02, 0.05
D) 0.02, 0.15

Answer 1

Hint: To find the gram atom we have to find the number of moles of respective component. Mole is the unit of measurement for the amount of substance.
At first, we have to find the atomic weight of Al, ${\text{C}}{{\text{l}}_{\text{2}}}$ and $\operatorname{Al}{\text{C}}{l_3}$.
Then, by dividing the values from grams we can find the mole. In the given problem given to solve the number of atoms in gram i.e. we have to find the mole of the respective component.

Complete step by step answer:
We can write the reaction as-
${\text{Al + C}}{{\text{l}}_2}{\text{ }} \to {\text{ AlC}}{{\text{l}}_3} \\
{\text{2Al + 3C}}{{\text{l}}_2}{\text{ }}\to {\text{ 2AlC}}{{\text{l}}_3} \\$
Atomic weight of Al = 27 g
Atomic weight of $Cl_2$ = 35.5 g
So, we can say molecular weight of ${\text{C}}{{\text{l}}_{\text{2}}}\; = \;2 \times 35.5\; = \;71g$

\[\therefore {\text{1}}\;{\text{mole of C}}{{\text{l}}_2}{\text{ = 71g C}}{{\text{l}}_2}\]

Molecular weight of $AlCl_3$ = $27 + 3 \times 35.5$
$ = 27 + 106.5$
= 133.5 g
\[\therefore {\text{1}}\;{\text{mole of AlC}}{{\text{l}}_3}{\text{ = 133}}{\text{.5 g AlC}}{{\text{l}}_3}\]
Moles of $\operatorname{Al}{\text{C}}{l_3}$ produced $ = \dfrac{6.67 g}{133.5}$ = 0.05mol
Now, 0.54 g Al remains unreacted.
${\text{27 g}}\;{\text{of}}\;{\text{Al}}\;{\text{ = }}\;{\text{1}}\;{\text{mole}}$
$0.54\;g\;of\;Al\; = \dfrac{{0.54}}{{27}}\; = \;0.02mol{\text{e}}$
So gram atoms or moles of Al taken $ = \;0.05 + 0.02\; = \;0.07mol{\text{e}}$
Gram atoms or moles of ${\text{C}}{{\text{l}}_{\text{2}}}\;taken\; = \;3\times0.05\; = \;0.15{\text{ mole}}$

$\therefore$ option (A) is correct answer.

Note: To find this type of problem we have to remember that at first, we have to find the atomic weight, molecular weight and mole of the given atom or compound.
We need to be careful while converting grams to moles.
We need to take care while determining the molecular weight of monatomic or polyatomic molecules.
We need to remember that gram atoms refer to moles, i.e. the number of gram atoms is equal to number of mole.