An alloy of copper and aluminum was treated with an excess of sodium hydroxide solution, while a 1.344

An alloy of copper and aluminum was treated with an excess of sodium hydroxide solution, while a 1.344 liter gas was released. The residue after the reaction was dissolved in nitric acid, then the solution was evaporated and calcined to constant weight, which turned out to be equal to 0.4 g. What is the composition of the alloy?

Given:
V (gas) = 1.344 l
m (residue) = 0.4 g

To find:
ω (Cu) -?
ω (Al) -?

Decision:
1) 2Al + 2NaOH + 6H2O => 2Na [Al (OH) 4] + 3H2 ↑;
3Cu + 8HNO3 => 3Cu (NO3) 2 + 2NO + 4H2O;
2Cu (NO3) 2 => 2CuO + 4NO2 + O2;
2) n (H2) = V (H2) / Vm = 1.344 / 22.4 = 0.06 mol;
3) n (Al) = n (H2) * 2/3 = 0.06 * 2/3 = 0.04 mol;
4) m (Al) = n (Al) * M (Al) = 0.04 * 27 = 1.08 g;
5) n (CuO) = m (CuO) / M (CuO) = 0.4 / 80 = 0.005 mol;
6) n (Cu) = n (Cu (NO3) 2) = n (CuO) = 0.005 mol;
7) m (Cu) = n (Cu) * M (Cu) = 0.005 * 64 = 0.32 g;
8) m alloy (Cu, Al) = m (Cu) + m (Al) = 0.32 + 1.08 = 1.4;
9) ω (Cu) = m (Cu) * 100% / m alloy (Cu, Al) = 0.32 * 100% / 1.4 = 22.857%;
10) ω (Al) = 100% – ω (Cu) = 100% – 22.857% = 77.143%.

Answer: Mass fraction of Cu is 22.857%; Al – 77.143%.