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.