An alloy of calcium with an unknown metal weighing 13.2 g was dissolved in hydrochloric acid. As a result of the reaction

An alloy of calcium with an unknown metal weighing 13.2 g was dissolved in hydrochloric acid. As a result of the reaction, 6.72 liters of hydrogen was released. Identify the unknown metal. It is known that the mass fraction of calcium in the alloy is 60.6%.

Given:
m alloy (Ca, Me) = 13.2 g
V (H2) = 6.72 L
ω (Ca) = 60.6%

To find:
Me -?

Decision:
1) Ca + 2HCl => CaCl2 + H2 ↑;
2) m (Ca) = ω (Ca) * m alloy (Ca, Me) / 100% = 60.6% * 13.2 / 100% = 8 g;
3) n (Ca) = m (Ca) / M (Ca) = 8/40 = 0.2 mol;
4) n1 (H2) = n (Ca) = 0.2 mol;
5) V1 (H2) = n1 (H2) * Vm = 0.2 * 22.4 = 4.48 l;
6) V2 (H2) = V (H2) – V1 (H2) = 6.72 – 4.48 = 2.24 l;
7) Me + HCl -> MeClх + H2 ↑;
8) m (Me) = m alloy (Ca, Me) – m (Ca) = 13.2 – 8 = 5.2 g;
9) m (Me) / M eq. (Me) = V2 (H2) / V eq. (H2);
M equiv. (Me) = m (Me) * V eq. (H2) / V2 (H2) = 5.2 * 11.2 / 2.24 = 26 g / mol;
10) z = 1, Ar (Me) = 26 – there is no such Me;
z = 2, Ar (Me) = 52 – Cr – chromium.

Answer: Unknown metal – Cr – chromium.