Tinned metal weighing 18.4 g was treated with water and 8.96 liters of gas were released, what kind of metal is it?

| February 2, 2021 | education

Let M be an unknown metal. Since the condition does not say anything about the valency of the metal, we will take it equal to one. The gas released during the interaction of M with water is hydrogen with a high probability (it is known that metals in the electrochemical range of activity to the left of hydrogen are capable of displacing hydrogen from water and acids).

Let’s write the reaction equation:

2M + 2H2O = 2MOH + H2 ↑

The amount of hydrogen substance:

v (H2) = V (H2) / Vm = 8.96 / 22.4 = 0.4 (mol).

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

v (M) = v (H2) * 2 = 0.4 * 2 = 0.8 (mol).

Thus, the molar mass of the unknown metal M:

M (M) = m (M) / v (M) = 18.4 / 0.8 = 23 (g / mol), which corresponds to the metal sodium (Na).



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