An alloy of magnesium with copper weighing 27 g was dissolved in an excess of chloride acid. In this case, 16.8 liters

An alloy of magnesium with copper weighing 27 g was dissolved in an excess of chloride acid. In this case, 16.8 liters of hydrogen were released. Determine the mass fractions of metals in the alloy.

Given:
m alloy (Mg, Cu) = 27 g
V (H2) = 16.8 L

To find:
ω (Mg) -?
ω (Cu) -?

Decision:
1) Mg + 2HCl => MgCl2 + H2 ↑;
Cu + HCl – the reaction does not proceed;
2) M (Mg) = Mr (Mg) = Ar (Mg) = 24 g / mol;
3) n (H2) = V (H2) / Vm = 16.8 / 22.4 = 0.75 mol;
4) n (Mg) = n (H2) = 0.75 mol;
5) m (Mg) = n (Mg) * M (Mg) = 0.75 * 24 = 18 g;
6) ω (Mg) = m (Mg) * 100% / m alloy (Mg, Cu) = 18 * 100% / 27 = 66.67%;
7) ω (Cu) = 100% – ω (Mg) = 100% – 66.67% = 33.33%.

Answer: The mass fraction of Mg is 66.67%; Cu – 33.33%.