An alloy of zinc and copper with a mass of 20 g was treated with an excess of sodium hydroxide solution with a mass

An alloy of zinc and copper with a mass of 20 g was treated with an excess of sodium hydroxide solution with a mass fraction of 30% (p = 1.33 g / ml). The solid residue was isolated and treated with an excess of concentrated nitric acid solution. The salt formed in this process was idled and calcined. The mass of the solid sediment composition is 10.016 g. Calculate the mass fractions of metals in the alloy, the consumed volume of alkali solution.

Given:
m alloy (Zn, Cu) = 20 g
ω (NaOH) = 30%
ρ solution (NaOH) = 1.33 g / ml
m (residue) = 10.016 g

To find:
ω (Zn) -?
ω (Cu) -?
V solution (NaOH) -?

Decision:
1) Zn + 2NaOH + 2H2O => Na2 [Zn (OH) 4] + H2;
Cu + 4HNO3 => Cu (NO3) 2 + 2NO2 + 2H2O;
2Cu (NO3) 2 => 2CuO + 4NO2 + O2;
2) n (CuO) = m (CuO) / M (CuO) = 10.016 / 80 = 0.1252 mol;
3) n (Cu) = n (Cu (NO3) 2) = n (CuO) = 0.1252 mol;
4) m (Cu) = n (Cu) * M (Cu) = 0.1252 * 64 = 8.0128 g;
5) ω (Cu) = m (Cu) * 100% / m alloy (Zn, Cu) = 8.0128 * 100% / 20 = 40.064%;
6) ω (Zn) = 100% – ω (Cu) = 100% – 40.064% = 59.936%;
7) m (Zn) = m alloy (Zn, Cu) – m (Cu) = 20 – 8.0128 = 11.9872 g;
8) n (Zn) = m (Zn) / M (Zn) = 11.9872 / 65 = 0.1844 mol;
9) n (NaOH) = n (Zn) * 2 = 0.1844 * 2 = 0.3688 mol;
10) m (NaOH) = n (NaOH) * M (NaOH) = 0.3688 * 40 = 14.752 g;
11) m solution (NaOH) = m (NaOH) * 100% / ω (NaOH) = 14.752 * 100% / 30% = 49.1733 g;
12) V solution (NaOH) = m solution (NaOH) / ρ solution (NaOH) = 49.1733 / 1.33 = 36.972 ml.

Answer: The mass fraction of Zn is 59.936%; Cu – 40.064%; the volume of NaOH is 36.972 ml.