An alloy of Fe and Cu weighing 18.4 g was treated in the cold with concentrated nitric acid, while 8.96 liters

An alloy of Fe and Cu weighing 18.4 g was treated in the cold with concentrated nitric acid, while 8.96 liters of gas were released. Calculate the total number of atoms in the original mixture.

Given:
m alloy (Fe, Cu) = 18.4 g
V (gas) = 8.96 l

To find:
N (Fe, Cu) -?

Solution:
1) Cu + 4HNO3 => Cu (NO3) 2 + 2NO2 ↑ + 2H2O;
Fe + HNO3 – passivation;
2) n (NO2) = V (NO2) / Vm = 8.96 / 22.4 = 0.4 mol;
3) n (Cu) = n (NO2) / 2 = 0.4 / 2 = 0.2 mol;
4) N (Cu) = n (Cu) * NA = 0.2 * 6.02 * 1023 = 1.204 * 10 ^ 23;
5) m (Cu) = n (Cu) * M (Cu) = 0.2 * 64 = 12.8 g;
6) m (Fe) = m alloy (Fe, Cu) – m (Cu) = 18.4 – 12.8 = 5.6 g;
7) n (Fe) = m (Fe) / M (Fe) = 5.6 / 56 = 0.1 mol;
8) N (Fe) = n (Fe) * NA = 0.1 * 6.02 * 10 ^ 23 = 0.602 * 10 ^ 23;
9) N (Fe, Cu) = N (Cu) + N (Fe) = 1.204 * 10 ^ 23 + 0.602 * 10 ^ 23 = 1.806 * 10 ^ 23.

Answer: The total number of Fe and Cu atoms is 1.806 * 10 ^ 23.