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 $Cl_2$.

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

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

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 $\frac{0.324}{3}\times 2=0.216$ 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 $\frac{0.324}{3}\times 2=0.216$ moles of $AlCl_3$

Now we have to calculate the mass of $AlCl_3$

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

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

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

Answer 2

Number of moles of aluminum chloride = 0.216. 
Mass of compound = Number of moles x Molar Mass 
= 0.216mol x [ 37 + 3(35.5) ]g/mol 
= 30.996 
= 31.0g (3 s.f.)