A piece of silver-nickel alloy weighing 1.5 g was placed in a glass with hydrochloric acid, after 0.2 l

A piece of silver-nickel alloy weighing 1.5 g was placed in a glass with hydrochloric acid, after 0.2 l was released. hydrogen and the reaction stopped, the rest of the alloy was removed from the solution. What metal was left, what is its mass?

Given:
m (alloy) = 1.5 g
V (H2) = 0.2 l

To find:
Me -?
m (Me) -?

1) 2HCl + Ni => NiCl2 + H2 ↑;
2) Silver remains in the rest of the alloy, which does not interact with HCl, since it is located to the right of H in the electrochemical range of metal activity;
2) n (H2) = V (H2) / Vm = 0.2 / 22.4 = 0.01 mol;
3) n (Ni) = n (H2) = 0.01 mol;
4) M (Ni) = Ar (Ni) = 59 g / mol;
5) m (Ni) = n (Ni) * M (Ni) = 0.01 * 59 = 0.59 g;
6) m (Ag) = m (alloy) – m (Ni) = 1.5 – 0.59 = 0.91 g.

Answer: In the rest of the alloy – Ag; weight – 0.91 g.