Question

2()+32()⟶232Al(s)+3Cl2(g)⟶2AlCl3If you react 5.9 mol of Cl2 how many moles of AlCl3 is produced?

Answer 1

We have the following reaction:

`2Al+3Cl_2\rightarrow2AlCl_3`

So as we can see for every 3 moles of Cl2 we get 2 moles of AlCl3. So to calculate how many moles of AlCl3 we get with 5.9 moles of Cl2 we do the following:

`moles_(AlCl3)=5.9\text{ }mol_(Cl2)\text{ }x\text{ }\frac{2\text{ }mol_(AlCl3)}{3\text{ }mol_(Cl2)}=3.933\text{ }mol_(AlCl3)`

So the answer is 3.933 moles of AlCl3