Answer to Question #75551 in General Chemistry for Lilit

Question

Many metals react with halogens to give metal halides. For example,
2 Al (s) + 3 Cl2(g) → 2 AlCl3 (s)
If you begin with 13.5 g of aluminum

Answer 1

To find mass of aluminum chloride, we should find moles of aluminum. Moles of aluminum chloride AlCl3 are equal moles of aluminum, because they both have the same coefficient it the reaction:
2Al(s) + 3Cl2(g) → 2AlCl3(s)
n(AlCl_3 )=n(Al)=(m(Al))/(M(Al))=(13.5 g)/(26.98 g/mol)=0.5 mol
m(AlCl_3 )=n(AlCl_3 )×M(AlCl_3 )=0.5 mol×133.34 g/mol=66.7 g
Answer:
If you begin with 13.5 g of aluminum, you can get 66.7 g of AlCl3.