Determine the theoretical yield of h2s (in moles) if 32 molal2s3 and 32 molh2o are reacted according to the following balanced reaction. a possibly useful molar mass is al2s3 = 150.17 g/mol. al2s3(s) + 6 h2o(l) → 2 al(oh)3(s) + 3 h2s(g)

1) By looking at the balanced equation, you can see that 1 mol of Al2S3 needs 6 moles of H2O to fully react . So to react 32 moles, you need 6x32moles of water.
  moles H2O needed = 192 moles H2O
However, you don't have 192 moles, you only have 32- This means the water is the limiting reagent.

So, to fully react 32 moles of water you only need 1/6 of Al2S3

moles of Al2S3 needed= 32* 1/6   =5,33 mol

2)
For the theoretical yield to be maximum amount of product possible we need the limiting reagent  to fully react.

So how many moles of H2S we obtain, from the 32 moles of H2O available to react?

6 moles H2O react to produce 3 moles H2S
moles H2S = 32 x (3 mol H2S / 6 mol H2O)
= 16 mol H2S

the theoretical yield of H2S = 16 mol

The correct name for n3-ion should be azide.