Question

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl2(g)→2AlCl3(s)

You are given 10.0 g of aluminum and 15.0 g of chlorine gas.

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 10.0 g of aluminum?

Answer 1

moles of Al = mass / molar mass = 10 / 27 = 0.37

2 Al(s)+ 3 Cl2(g) ------------------->   2 AlCl3(s)

2                                                     2

0.37                                                   ??

2 moles of Al ---------------- > gives 2 mol of AlCl3

0.37 moles of Al ---------------> ??

moles of AlCl3 = 0.37 moles