Question

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

Answer 1

Answer: 0.70 moles

Explanation:

$2Al+3Cl_2\rightarrow 2AlCl_3$

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 =$\frac{\text{ given mass of Al}}{\text{ molar mass of Al}}= \frac{19g}{27g/mole}=0.70moles$

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 =$\frac{2}{2}\times 0.70=0.70moles$ of aluminium chloride.

Thus 0.70 moles of aluminium chloride will be formed.

Answer 2

The chemical reaction would be written as follows:

2Al + 3Cl2 = 2AlCl3

We are given the amount of aluminum to be used in the reaction. This will be the starting point of the calculations. We do as follows:

19.0 g Al ( 1 mol / 29.98 g ) ( 2 mol AlCl3 / 2 mol Al ) = 0.63 mol AlCl3