270 g of aluminum was dissolved in hydrochloric acid, containing 10% of impurities. What volume of hydrogen was obtained

270 g of aluminum was dissolved in hydrochloric acid, containing 10% of impurities. What volume of hydrogen was obtained in this case if its yield is 75% of the theoretically possible

1. The reaction proceeds according to the equation:

2Al + 6HCl = 2AlCl3 + 3H2 ↑;

2. Let’s calculate the mass of aluminum:

m (Al) = m (sample) – m (impurities);

m (impurities) = w (impurities) * m (sample) = 0.1 * 270 = 27 g;

m (Al) = 270 – 27 = 243 g;

3. Let’s calculate the chemical amount of aluminum:

n (Al) = m (Al): M (Al) = 243: 27 = 9 mol;

4. Determine the theoretical amount of hydrogen:

ntheor (H2) = n (Al) * 3: 2 = 9 * 3: 2 = 13.5 mol;

5. Set the practical amount of hydrogen:

npr (H2) = ν * ntheor (H2) = 0.75 * 13.5 = 10.125 mol;

6. Find the volume of the obtained gas:

V (H2) = npr (H2) * Vm = 10.125 * 22.4 = 226.8 liters.

Answer: 226.8 liters.