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}=(20.0g)/(27g/mol)=0.741mol`

and,

`\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}=(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
`(2)/(3)* 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
`(2)/(3)* 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* \text{ Molar mass of }AlCl_3`

Molar mass of
`AlCl_3` = 133 g/mole

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

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