Question

Determine the theoretical yield of H2S (in moles) if 4.0 molAl2S3 and 4.0 mol H2O are reacted according to the followingbalanced reaction. A possibly useful molar mass is Al2S3 =150.17Al2S3(s)+6H2O(l)-->2Al(OH)3(s)+3H2S(g)

Answer 1

Answer:

The theoretical yield of the hydrogen sulfide is 68.0 grams.

Explanation:

$Al_2S_3(s)+6H_2O(l)\rightarrow 2Al(OH)_3(s)+3H_2S(g)$

Moles of aluminum sulfide = 4.0 mol

Moles of water = 4.0 mol

According to reaction, 6 moles of water reacts with 1 mole of aluminum sulfide,then 4 moles of water will react with :

$\frac{1}{6}\times 4 mol=1.5 mol$ of aluminum sulfide

1.5 moles aluminum sulfide < 4 moles  aluminum sulfide

This means that water is present in limiting amount and aluminum sulfide is in excess amount.So, amount of hydrogen sulfide will depend upon moles of water.

According to reaction, 6 moles of water gives with 3 mole of hydrogen sulfide,then 4 moles of water will give :

$\frac{3}{6}\times 4 mol=2.0 mol$ of hydrogen sulfide

Mass of hydrogen sulfide:

2.0 mol × 34 g/mol = 68.0 g

The theoretical yield of the hydrogen sulfide is 68.0 grams.