Question #127634
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?
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?
Expert's answer
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
