Question

Use this equation- 2 Al + 3 F2 -> 2 AlF3
1. If 10.0 moles of aluminum react with 10.0 moles of fluorine gas, how many moles of aluminum fluoride will be produced?
Explain how you did it pls.

Answer 1

The equation is:  2 Al + 3 F2 ----> 2 AlF3

Assuming that, 10 moles of Aluminium is reacting completely with the required amount of fluorine gas to produce aluminium fluoride.

According to the balanced chemical equation, 2 parts of Aluminium react with 3 parts of fluorine to produce 2 parts of aluminium fluoride.

So, 2 parts is 10 mol, 1 part is hence 5 mol.

So, 3 parts of fluorine gas will be 3*5 => 15 mol , hence for the amount of fluorine available, the reaction will not be completed for 10 moles of aluminium as only 10 moles of fluorine gas are available.

Now, assuming that the 10 moles of fluorine react completely with the required amount of aluminium metal to produce aluminium fluoride.

So, from the balanced chemical equation, 10 moles of fluorine is 3 parts.

So, 1 part will be 10/3 moles.

So, 2*10/3 =>  20/3 moles of aluminium will be required (6.67 moles of Al), which is easily available.

Hence, 2 parts of aluminium fluoride or 2*10/3 moles => 20/3 moles of Aluminium fluoride will be produced or 6.67 mol of Aluminium fluoride will form.

Hence, 10-6.67 mol or 3.33 moles of aluminium will remain unused.

1 mole of aluminium has a mass of 27 grams.

So, 27*3.33 grams or 89.91 grams of aluminium will remain unused.