25.0 g of an alloy of magnesium and aluminum was treated with an excess of alkali.

25.0 g of an alloy of magnesium and aluminum was treated with an excess of alkali. The gas released in this case completely reduced 32.0 g of copper (II) oskil. Calculate the mass fractions (in%) of metals in the alloy.

Let’s write down the reaction equations:
Mg + NaOH = no reaction, only aluminum reacts with alkali.

2Al + 2NaOH + 6H2O = 2Na [Al (OH) 4] + 3H2 ↑

H2 + CuO = Cu + H2O

Let’s calculate the amount of hydrogen:
n (CuO) = m (CuO) / Mr (CuO) = 32/80 = 0.4 mol (the molar mass of copper is taken as 64 g / mol);

n (H2) = n (CuO) = 0.4 mol;

Let’s calculate the mass of aluminum in the alloy:
n (Al) = 2/3 * n (H2) = 0.8 / 3 = (approximately) 0.267 mol;

m (Al) = n (Al) * Mr (Al) = 0.267 * 27 = (approximately) 7.2 g;

Let’s find the mass fractions of metals in the alloy:
ω (Al) = m (Al) / m (alloy) * 100% = 7.2 / 25 * 100 = 28.8%

ω (Mg) = 100% – ω (Al) = 100 – 28.8 = 71.2%