As a result of reactions with water of a metal weighing 31.2 g, which is in group I, 8.96 gas was released. Identify the metal.

To solve the problem, let’s compose the equation schematically:
m = 31.2 g; V = 8.96 liters.
1.2Me + H2O = 2MeOH + H2 – OBP, hydrogen is released;
Determine the metal of the 1st group.
2. Let’s calculate the amount of H2:
1 mol of gas at normal level – 22.4 liters;
X mol (H2) – 8.96 L. hence, X mol (H2) = 1 * 8.96 / 22.4 = 0.4 mol.
3. Find the molar mass of the unknown metal:
Y (Me) = m / M; M = 31.2 / 0.4 = 78;
78/2 = 39;
Ar (K) = 39;
Answer: in the first group of the periodic system – potassium.