Answer to Question #127634 in General Chemistry for Shaharyar Salman

Question

The following chemical reaction takes place:
3 (NH4)2S(aq) + 2 AlCl3(aq) = Al2S3(s) + 6 NH3(g) + 6HCl(aq)
If a 400 mL solution of AlCl3 with a concentration of 1.10 mol/L is reacted with 3.56g (NH4)2S:
i) What is the concentration of HCl?
ii) How much Al2S3(s) is produced?
iii) If this reaction is carried at SATP conditions, what volume of NH3 is produced?

Answer 1

First the limiting reactant should be determined.

n (AlCl₃) = 1.10M x 0.4L = 0.44 mol

n (NH₄)₂S = 3.56g / 68.15g/mol = 0.0522 mol - evidently limiting reactant.

1) n (HCl) = 6/3 x n(NH₄)₂S = 0.1044 mol

[HCl] = 0.1044mol / 0.4L = 0.261 M

2) n (Al₂S₃) = 1/3 x n(NH4)2S = 0.0174 mol

m (Al₂S₃) = 0.0174mol x 150.16g/mol = 2.61 g

3) n (NH₃) = 6/3 x n(NH₄)₂S = 0.1044 mol

V (NH₃) = 0.1044mol x 22.4L/mol = 2.34 L