Upon the interaction of 6.85 g of metal with water, 1.12 liters of hydrogen were released

Upon the interaction of 6.85 g of metal with water, 1.12 liters of hydrogen were released, determine this metal if it is divalent in its compounds.

The reaction of interaction of an unknown metal Me (II) with water looks like this:
Me + H2O = MeO + H2.
The amount of hydrogen produced during the reaction:
1.12: 22.4 = 0.05 mol, from which we get the molar mass of the metal:
6.85 g: 0.05 mol = 137 g / mol. This is barium (Ba).