Question

calculate the grams of indicated product when 16.6 g

g of the first reactant and 10.4 g

g of the second reactant is used: Al

2

S

3

(s)+6H

2

O(l)→2Al(OH)

3

(aq)+3H

2

S(g)

Al2S3(s)+6H2O(l)→2Al(OH)3(aq)+3H2S(g) (H

2

S)

Answer 1

$Al2S3(s)+6H2O(l)→2Al(OH)3(aq)+3H2S(g) (H 2 S)$

Moles of $Al_2S_3=\frac{16.6}{150.158 }=0.11$

Moles $H_2O=\frac{10.4}{18.02}=0.577moles$