An alloy of silver and copper weighing 27.93 g was dissolved in conc. nitric acid, the solution was evaporated

An alloy of silver and copper weighing 27.93 g was dissolved in conc. nitric acid, the solution was evaporated, and the residue was calcined. 12.32 liters of gas were released. What are the mass fractions of metals in the alloy.

Given:
m alloy (Ag, Cu) = 27.93 g
V (gas) = ​​12.32 l

To find:
ω (Ag) -?
ω (Cu) -?

Decision:
1) Ag + 2HNO3 => AgNO3 + NO2 + H2O;
Cu + 4HNO3 => Cu (NO3) 2 + 2NO2 + 2H2O;
2AgNO3 => 2Ag + 2NO2 + O2;
2Cu (NO3) 2 => 2CuO + 4NO2 + O2;
2) n (NO2, O2) = V (NO2, O2) / Vm = 12.32 / 22.4 = 0.55 mol;
3) Let n (Ag) = (x) mol, and n (Cu) = (y) mol;
4) n (AgNO3) = n (Ag) = (x) mol;
5) n1 (NO2) = n (AgNO3) = (x) mol;
6) n1 (O2) = n (AgNO3) / 2 = (x / 2) mol;
7) n (Cu (NO3) 2) = n (Cu) = (y) mol;
8) n2 (NO2) = n (Cu (NO3) 2) * 2 = (y * 2) mol;
9) n2 (O2) = n (Cu (NO3) 2) / 2 = (y / 2) mol;
10) 108x + 64y = 27.93;
x + (x / 2) + 2y + (y / 2) = 0.55;
y = 0.22 – 0.6x;
108x + 14.08 – 38.4x = 27.93;
x = 0.2;
11) m (Ag) = n (Ag) * M (Ag) = x * 108 = 0.2 * 108 = 21.6 g;
12) ω (Ag) = m (Ag) * 100% / m alloy = 21.6 * 100% / 27.93 = 77.34%;
13) ω (Cu) = 100% – ω (Ag) = 100% – 77.34% = 22.66%.

Answer: Mass fraction of Ag is 77.34%; Cu – 22.66%.