Question

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) What is the maximum mass of aluminum chloride that can be formed when reacting 32.0 g of aluminum with 37.0 g of chlorine? Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer: The mass of aluminium chloride that can be formed are 46.3 g

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  ....(1)  

• For Aluminium:

Given mass of aluminium = 32 g  

Molar mass of aluminium = 26.98 g/mol

Putting values in above equation, we get:  

$\text{Moles of aluminium}=\frac{32g}{26.98g/mol}=1.186mol$

• For Chlorine:

Given mass of chlorine = 37 g  

Molar mass of chlorine = 71 g/mol

Putting values in above equation, we get:  

$\text{Moles of chlorine gas}=\frac{37g}{71g/mol}=0.521mol$

For the given chemical equation:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)$

By Stoichiometry of the reaction:

3 moles of chlorine gas is reacting with 2 moles of aluminium.

So, 0.521 moles of chlorine gas will react with = $\frac{2}{3}\times 0.521=0.347moles$ of aluminium.

As, given amount of aluminium is more than the required amount. Thus, it is considered as an excess reagent.

So, chlorine gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

3 moles of chlorine gas is producing 2 moles of aluminium chloride

So,  0.521 moles of chlorine gas will react with = $\frac{2}{3}\times 0.521=0.347moles$ of aluminium chloride.

Now, calculating the mass of aluminium chloride by using equation 1, we get:

Moles of aluminium chloride = 0.347 moles

Molar mass of aluminium chloride = 133.34 g/mol

Putting all the values in equation 1, we get:

$0.347mol=\frac{\text{Mass of aluminium chloride}}{133.34g/mol}\\\\\text{Mass of aluminium chloride}=46.3g$

Hence, the mass of aluminium chloride that can be formed are 46.3 g