Sulfur weighing 16 g reacts with aluminum. the resulting salt was dissolved in water

Sulfur weighing 16 g reacts with aluminum. the resulting salt was dissolved in water, the resulting gas was burned. how much oxygen was required?

Let’s implement the solution:
1. According to the condition of the problem, we compose the equations:
m = 16 g;
3S + 2Al = Al2S3 – compounds, obtained aluminum sulfide;
Al2S3 + 6H2O = 2Al (OH) 3 + 3H2S – complete hydrolysis, hydrogen sulfide is released;
X l. -?
2H2S + 3O2 = 3SO2 + 2H2O – hydrogen sulfide combustion.
2. Calculations:
M (S) = 32 g / mol;
M (O2) = 32 g / mol;
Y (S) = m / M = 16/32 = 0.5 mol.
3. Proportions:
0.5 mol (S) – X mol (Al2S3);
– 3 mol – 1 mol from here, X mol (Al2S3) = 0.5 * 1/3 = 0.16 mol;
0.16 mol (Al2S3) – X mol (H2S);
– 1 mol – 3 mol from here, X mol (H2S) = 0.16 * 3/1 = 0.48 mol.
0.48 mol (H2S) – X mol (O2);
– 2 mol – 3 mol from here, X mol (O2) = 0.48 * 3/2 = 0.72 mol.
4. Find the volume of O2:
V (O2) = 0.72 * 22.4 = 16.13 liters.
Answer: you need oxygen with a volume of 16.13 liters.