If you had excess chlorine, how many moles of of aluminum chloride could be produced from 19.0 g of aluminum?

The number of moles of aluminum chloride that could be produced is 0.704 moles

First, we will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction between aluminum and chlorine is

2Al + 3Cl₂ → 2AlCl₃

This means 2 moles of aluminum (Al) will react with 3 moles of chlorine (Cl₂) to produce 2 moles of aluminum chloride (AlCl₃)

Now, to determine the number of moles of aluminum chloride that could be produced,

We will determine the number of moles of aluminum present in 19.0 g of Al

Using the formula

Mass of aluminum = 19.0 g

Molar mass of aluminum = 26.98 g/mol

∴ Number of moles of aluminum present =

Number of moles of aluminum present = 0.704 moles

Since,

2 moles of aluminum (Al) reacts to produce 2 moles of aluminum chloride (AlCl₃)

Then,

0.704 moles of aluminum (Al) will react to produce 0.704 moles of aluminum chloride (AlCl₃)

Hence, the number of moles of aluminum chloride that could be produced is 0.704 moles

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0.70 moles

Explanation:

As can be seen from the chemical equation, 2 moles of aluminium react with 3 moles of chlorine.

According to mole concept, 1 mole of every substance weighs equal to its molar mass.

Moles of Al =

As aluminium is the limiting reagent as it limits the formation of product and chlorine is the excess reagent as it is left unreacted.

2 moles of aluminium reacts to produce 2 moles of aluminium chloride.

0.70 of aluminium reacts to form = of aluminium chloride.

Thus 0.70 moles of aluminium chloride will be formed.