Question

b. If 20.0 grams of Aluminum and 30.0 grams of chlorine gasſare used, and how many
grams AlCl3 can theoretically be made (3 pts)?

Answer 1

Answer: The mass of $AlCl_3$ theoretically be made can be, 37.4 grams.

Explanation : Given,

Mass of $Al$ = 20.0 g

Mass of $Cl_2$ = 30.0 g

Molar mass of $Al$ = 27 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Al$ and $Cl_2$.

$\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{20.0g}{27g/mol}=0.741mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}=\frac{30.0g}{71g/mol}=0.422mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$2Al+3Cl_2\rightarrow 2AlCl_3$

From the balanced reaction we conclude that

As, 3 moles of $Cl_2$ react with 2 moles of $Al$

So, 0.422 moles of $Cl_2$ react with $\frac{2}{3}\times 0.422=0.281$ moles of $Al$

From this we conclude that, $Al$ is an excess reagent because the given moles are greater than the required moles and $Cl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AlCl_3$

From the reaction, we conclude that

As, 3 moles of $Cl_2$ react to give 2 moles of $AlCl_3$

So, 0.422 moles of $Cl_2$ react to give $\frac{2}{3}\times 0.422=0.281$ mole of $AlCl_3$

Now we have to calculate the mass of $AlCl_3$

$\text{ Mass of }AlCl_3=\text{ Moles of }AlCl_3\times \text{ Molar mass of }AlCl_3$

Molar mass of $AlCl_3$ = 133 g/mole

$\text{ Mass of }AlCl_3=(0.281moles)\times (133g/mole)=37.4g$

Therefore, the mass of $AlCl_3$ theoretically be made can be, 37.4 grams.