When 3.42 g of alkali metal interacted with water, 448 cm3 of hydrogen was formed. Which metal reacted.

Let’s write the reaction equation:

2M + 2H2O = 2MOH + H2 ↑, where M is an unknown alkali metal.

Amount of substance M:

v (M) = m (M) / M (M) = 3.42 / M (M) (mol).

Amount of substance H2:

v (H2) = V (H2) / Vm = 0.448 / 22.4 = 0.02 (mol).

According to the reaction equation, 1 mole of H2 is formed per 2 mol of M, therefore:

v (M) = v (H2) * 2,

3.42 / M (M) = 0.02 * 2,

M (M) = 85.5 (g / mol), which corresponds to rubidium metal (Rb).

Answer: rubidium (Rb).