How many grams of aluminum sulfide would form if 2.25 liters of a 1.50 molar aluminum chloride solution reacts with 2.00 liters of a 2.75 molar hydrosulfuric acid (h2s) solution?
2alcl3 + 3h2s al2s3 + 6hcl (2 points)
• 18.6 g
• 101 g
• 253 g
• 275 g

1) Calculate the number of mols of each reagent, from the molar concentrations.

2.25 liters of a 1.50 molar aluminum chloride solution

M = n / V => n = M*V = 2.5liters*1.50 mol/liter = 3.75 mol

 2.00 liters of a 2.75 molar hydrosulfuric acid (H2S) solution

n = 2.00 liter * 2.75 mol/liter = 5.5 mol

2AlCl3 + 3H2S -> Al2S3 + 6HCl

Theoretical ratio 3 ACl3 / 2 H2S
Actual ratio 3.75 / 5.5 = 0.68 > 2/3 => H2S is the limitan reagent

5.5 mol H2S * 1 mol Al2S3 /  3mol H2S = 1.833 mol Al2S3

Molar mass of Al2S3 = 2* 27 g/mol + 3*32 g/mol = 150 g/mol

Mass of Al2S3 = 150g/mol*1.833mol = 274.95 g

275 g

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