Respuesta :

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

Solution : Given,

Mass of Al = 18.0 g

Mass of [tex]Cl_2[/tex] = 23.0 g

Molar mass of Al = 27 g/mole

Molar mass of [tex]Cl_2[/tex] = 71 g/mole

Molar mass of [tex]AlCl_3[/tex] = 133.5 g/mole

First we have to calculate the moles of Al and [tex]Cl_2[/tex].

[tex]\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{18.0g}{27g/mole}=0.667moles[/tex]

[tex]\text{ Moles of }Cl_2=\frac{\text{ Mass of }Cl_2}{\text{ Molar mass of }Cl_2}=\frac{23.0g}{71g/mole}=0.324moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]2Al+3Cl_2\rightarrow 2AlCl_3[/tex]

From the balanced reaction we conclude that

As, 3 mole of [tex]Cl_2[/tex] react with 2 mole of [tex]Al[/tex]

So, 0.324 moles of [tex]Cl_2[/tex] react with [tex]\frac{0.324}{3}\times 2=0.216[/tex] moles of [tex]Al[/tex]

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

Now we have to calculate the moles of [tex]AlCl_3[/tex]

From the reaction, we conclude that

As, 3 mole of [tex]Cl_2[/tex] react to give 2 mole of [tex]AlCl_3[/tex]

So, 0.324 moles of [tex]Cl_2[/tex] react to give [tex]\frac{0.324}{3}\times 2=0.216[/tex] moles of [tex]AlCl_3[/tex]

Now we have to calculate the mass of [tex]AlCl_3[/tex]

[tex]\text{ Mass of }AlCl_3=\text{ Moles of }AlCl_3\times \text{ Molar mass of }AlCl_3[/tex]

[tex]\text{ Mass of }AlCl_3=(0.216moles)\times (133.5g/mole)=28.8g[/tex]

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