An alloy of manganese with copper, weighing 100 g, was treated with hydrochloric acid. In this case

An alloy of manganese with copper, weighing 100 g, was treated with hydrochloric acid. In this case, 29.12 liters of hydrogen were released. Determine the mass fraction of manganese in the alloy.

Copper does not displace hydrogen from acids, since it is located to the right of it in the electrochemical series of metal activity, in contrast to manganese.
The equation for the reaction of manganese with hydrochloric acid:
Mn + 2HCl = MnCl2 + H2
The amount of hydrogen evolved under normal conditions (n.a.):
v (H2) = V (H2) / Vm = 29.12 / 22.4 = 1.3 (mol).
According to the reaction equation, 1 mol of H2 is formed from 1 mol of Mn, therefore:
v (Mn) = v (H2) = 1.3 (mol).
Manganese weight in the alloy:
m (Mn) = v (Mn) * M (Mn) = 1.3 * 55 = 71.5 (g)
and the mass fraction of manganese in the alloy:
w (Mn) = (m (Mn) / m (alloy)) * 100 = (71.5 / 100) * 100 = 71.5 (%).