Aluminum reacts with 49 g of sulfate acid. Calculate: 1) the mass of the metal that has reacted. 2) the volume of gas that was released.

1. We write down the equation:

2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2 ↑;

2.Calculate the chemical amount of acid:

n (H2SO4) = m (H2SO4): M (H2SO4);

M (H2SO4) = 2 + 32 + 4 * 16 = 98 g / mol;

n (H2SO4) = 49: 98 = 0.5 mol;

3. Determine the amount of aluminum and hydrogen:

n (Al) = n (H2SO4) * 2: 3 = 0.5 * 2: 3 = 0.3333 mol;

n (H2) = n (H2SO4) = 0.5 mol;

4. find the mass of aluminum:

m (Al) = n (Al) * M (Al) = 0.3333 * 27 = 9 g;

5. calculate the volume of hydrogen:

V (H2) = n (H2) * Vm = 0.5 * 22.4 = 11.2 liters.

Answer: 9 g, 11.2 liters.