Question

What is the maximum mass of aluminum chloride that can be formed when reacting 18.0 g of aluminum with 23.0 g of chlorine?

Answer 1

Answer : The maximum mass of aluminium chloride formed can be 28.8 grams.

Solution : Given,

Mass of Al = 18.0 g

Mass of
`Cl_2` = 23.0 g

Molar mass of Al = 27 g/mole

Molar mass of
`Cl_2` = 71 g/mole

Molar mass of
`AlCl_3` = 133.5 g/mole

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

`\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=(18.0g)/(27g/mole)=0.667moles`

`\text{ Moles of }Cl_2=\frac{\text{ Mass of }Cl_2}{\text{ Molar mass of }Cl_2}=(23.0g)/(71g/mole)=0.324moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`2Al+3Cl_2\rightarrow 2AlCl_3`

From the balanced reaction we conclude that

As, 3 mole of
`Cl_2` react with 2 mole of
`Al`

So, 0.324 moles of
`Cl_2` react with
`(0.324)/(3)* 2=0.216` 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 mole of
`Cl_2` react to give 2 mole of
`AlCl_3`

So, 0.324 moles of
`Cl_2` react to give
`(0.324)/(3)* 2=0.216` moles 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`

`\text{ Mass of }AlCl_3=(0.216moles)* (133.5g/mole)=28.8g`

Therefore, the maximum mass of aluminium chloride formed can be 28.8 grams.