Question

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

Answer 1

Answer:

                0.312 mol of S

Solution:

              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

Result:

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

Answer 2

Given, moles of Al = 0.208

Balanced equation for the given chemical reaction:

2Al + 3S --> Al₂S₃  

The molar ratio between Al and S is 2:3

0.208 moles of Al x (3 moles of S / 2 moles of Al)

= 0.312 moles of S

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