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

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.