Answer to Question #64824 in General Chemistry for Maddy

Question

Calculate the grams of indicated product when 17.5 g of the first reactant and 10.1 g of the second reactant is used: Al2S3(s)+6H2O(l)→2Al(OH)3(aq)+3H2S(g)

Answer 1

Answer
n(Al₂S₃)=m(Al₂S₃)/M(Al₂S₃)=1,75g/150g/Ml=0,012 Ml
n(H₂O)=m(H₂O)/M(H₂O)=10,1g/18g/Ml=0,561Ml
1Mol Al₂S₃ : 6Mol H₂O for equation 0,012 Mol Al₂S₃ : 0,561 Mol H₂O
Or
1 Mol Al₂S₃ : 46,75 Mol H₂O
Al₂S₃ - shortcoming
H₂O – excess
Al₂S₃+6H₂O -> 2Al(OH)₃+3H₂S
1Mol Al₂S₃/0,012Mol=2 Mol Al(OH)₃/x Mol
n(Al(OH)₃)=0,024 Mol
m(Al(OH)₃)=n*M=1,872g
1 Mol Al₂S₃/0,012Mol=3Mol H₂S/x Mol
n(H₂S)=0,036 Mol
m(H₂S)=n*M=1,224 g