Question

In an experiment, $6.67g$ of $AlC{l_3}$ was produced and $0.54g$ of aluminium remained unreacted. How many gram or atoms of aluminum and chlorine were taken originally? $\left( {Al = 27,Cl = 35.5} \right)$
A) $0.07$,$0.15$
B) $0.07$,$0.05$
C) $0.02$,$0.05$
D) $0.02$,$0.15$

Answer 1

We know that the amount of moles in a given amount of any substance is equal to the grams of the substance divided by its molar weight.
The mole of the substance can be calculated by using the formula as,
${\text{Mole}} = \dfrac{{{\text{Weight of the substance}}}}{{{\text{Molecular weight}}}}$

Complete step by step answer:
We can write the chemical equation for this as,
$Al + 3/2C{l_2}{\text{ }}\xrightarrow{{}}AlC{l_3}$
From the above reaction, one mole of aluminium produces one mole aluminum trichloride.
We know that the molecular weight of aluminium trichloride is $133.5g/mol$
Now, calculate the number of moles of aluminum trichloride produced in the reaction using the formula for mole calculation.
Moles of a $AlC{l_3}$ produced $= \dfrac{{6.67}}{{133.5}} = 0.05mol$
Given,
The excess aluminum trichloride remaining in the reaction is $0.54g$.
The excess of aluminium in the reaction can be calculated as,
Excess of aluminum $= \dfrac{{0.54}}{{27}} = 0.02moles$
The total amount of aluminum taken is $\left( {0.05 + 0.02} \right) = 0.07$
The total amount of chlorine taken is $\left( {3 \times 0.05} \right) = 0.15$
Thus the gram or atoms of aluminum and chlorine were taken originally are $0.07\& 0.15$.

Therefore, the option A is correct.

Note:
Mole ratio:
A mole ratio is a ratio between the numbers of moles of any two species involved in a chemical reaction.
Example,
In the reaction ${\text{2}}{{\text{H}}_{\text{2}}}{\text{ + }}{{\text{O}}_{\text{2}}}\xrightarrow{{}}{\text{2}}{{\text{H}}_{\text{2}}}{\text{O}}$, the mole ratio can be written as $\dfrac{{{\text{2mol}}{{\text{H}}_{\text{2}}}}}{{{\text{1mol}}{{\text{O}}_{\text{2}}}}}.$
Find the number moles of ${\text{CaS}}$ are produced from ${\text{2}}{\text{.5molCaO}}$.
Given,
The number of moles of ${\text{CaO}}$ is ${\text{2}}{\text{.5mol}}{\text{.}}$
The balanced reaction is,
${\text{4HgS + 4CaO}}\xrightarrow{{}}{\text{4Hg + 3CaS + CaS}}{{\text{O}}_{\text{4}}}$
In the mole ratio, the coefficients of the balanced equation are used. Therefore the mole ratio is $\dfrac{{{\text{3molCaS}}}}{{{\text{4molCaO}}}}$.
The number of moles can be calculated as,
${\text{2}}{\text{.5molCaO}}\left( {\dfrac{{{\text{3molCaS}}}}{{{\text{4molCaO}}}}} \right){\text{ = }}\,{\text{1}}{\text{.875molCaS}}$
The number moles of ${\text{CaS}}$ are produced from ${\text{2}}{\text{.5molCaO}}$ is $1.875{\text{mol}}$.