What is the mass and volume of hydrogen formed when 1.00 g of an alloy of the following composition

What is the mass and volume of hydrogen formed when 1.00 g of an alloy of the following composition is dissolved in acid: 5% aluminum, 2% zinc, 93% magnesium?

In order to solve this problem, you first need to write the following reaction equations:
2Al + 6HCl = 2AlCl3 + 3H2
Zn + 2HCl = ZnCl2 + H2
Mg + 2HCl = MgCl2 + H2
then we calculate by the following formulas the masses of aluminum, zinc and magnesium.
m (Al) = m alloy × w = 1 × 0.05 = 0.05 g
m (Zn) = m alloy × w = 1 × 0.02 = 0.02 g
m (Mg) = m alloy × w = 1 × 0.93 = 0.93 g
Next, we calculate the amount of substance for aluminum, zinc and magnesium
n (Al) = m / M = 0.05 / 27 = 0.00185 mol
n (Zn) = m / M = 0.02 / 65 = 0.0003077 mol
n (Mg) = m / M = 0.93 / 24 = 0.03875 mol
Then we count the amount of substance of all hydrogen compounds
n (H2) = 0.00185 + 0.0003077 + 0.03875 = 0.0418354 mol
Knowing the amount of matter, we find the mass of hydrogen
m (H2) = n × M = 0.0418354 × 2 = 0.0836709 g
Next, we calculate the volume of hydrogen
V (H2) = n × Vm = 0.0418354 × 22.4 = 0.9371147 l
Answer: m (H2) = 0.0836709 g, V = 0.9371147 l.