As a result of the interaction of an alkali metal weighing 2.76 g with an excess of water

As a result of the interaction of an alkali metal weighing 2.76 g with an excess of water, a gas with a volume of 1, 344 liters was formed. Determine the metal.

In accordance with the condition of the problem, we write down the reaction process schematically, taking into account that the metal is alkaline, its valence is 1;
2Ме + 2Н2О = Н2 + 2Ме (ОН) – ОВР;
Determine the number of moles of hydrogen:
1 mol – 22.4 L;
X mol (H2) – 1.344 hence, X mol (H2) = 1 * 1.344 / 22.4 = 0.06 mol;
Let’s make the proportion:
X mol (Me) – 0.06 mol (H2);
-2 mol – 1 mol from here, X mol (Me) = 2 * 0.06 / 1 = 0.12 mol;
Let’s calculate the molar mass of the metal by the formula:
Y = m / M; M = m / Y;
M (Me) = 2.76 / 0.12 = 23 g / mol;
According to tabular data, Ar (Na) = 22.9 = 23;
Answer: the alkali metal sodium is involved in the reaction.