Question

In the following chemical reaction, how many grams of the excess reactant are left over when 4.21E-1 moles of Al(s) react with 1.39E+0 moles of Cl(g)?

2Al(s) + 3Cl(g) → 2AlCl(s)

Answer 1

there must be 2 atoms $Al$ for every 3 molecules $Cl_2$

Hence Chlorine $Cl_2$ ​ is the excess reactant and $Al$ the limiting reactant

Mass of $Al=0.421×27=11.367grams$

Mass of $Cl_2=1.39×71=98.69grams$

Left over $=98.69-11.367=87.323g$