How many moles of s are needed to combine with 0.208 mol al to give al2s3?

                0.312 mol of S

              The Balance Chemical equation is as follow,

                                        2 Al + 3 S   →   Al₂S₃

According to this equation,

                        2 moles of Al required  =  3 moles of S

Then,

                   0.208 moles of Al will require  =  X moles of S

Solving for X,

                      X   =  (0.208 mol × 3 mol) ÷ 2 mol

                     X  = 0.312 mol of S

0.312 moles of S are needed to combine with 0.208 mol Al to give Al₂S₃.

it's filtration, but this is an assumption so my apologies if i got this

sure im new too!

explanation: