Find the volume of hydrogen n.on which is formed by the interaction of 420 g of Mg with H3PO4.

Given:
m (Mg) = 420 g

Find:
V (H2) -?

1) Write the equation of a chemical reaction:
3Mg + 2H3PO4 => Mg3 (PO4) 2 + 3H2 ↑;
2) Determine the molar mass of Mg:
M (Mg) = Ar (Mg) = 24 g / mol;
3) Calculate the amount of Mg substance:
n (Mg) = m (Mg) / M (Mg) = 420/24 = 17.5 mol;
4) Determine the amount of substance H2 (taking into account the coefficients in the reaction equation):
n (H2) = n (Mg) = 17.5 mol;
5) Calculate the volume of H2 (under normal conditions):
V (H2) = n (H2) * Vm = 17.5 * 22.4 = 392 HP

Answer: The volume of H2 is 392 liters.