When 27.4 g of divalent metal interacted with water, 4.48 liters of hydrogen were released and the hydroxide

When 27.4 g of divalent metal interacted with water, 4.48 liters of hydrogen were released and the hydroxide of this metal was formed. Identify the metal.

First, we write down the reaction equations.

Me + 2HOH = Me (OH) 2 + H2.

Let’s determine the mass of hydrogen.

n = m / M.

n = V / Vm.

m / M = V / Vm.

Let’s express the mass.

m = M * V / Vm.

M (H2) = 1 * 2 = 2 g / mol.

m = 2 * 4.48 / 22.4 = 0.4 g.

Let us determine the molar mass of the metal by the reaction equation.

27.4 g of metal – 0.4 g of hydrogen.

X g / mol of metal – 2 g / mol of hydrogen.

X = 2 * 27.4 / 0.4 = 137 g / mol.

Therefore, this metal has a molar mass of 137 g / mol. The divalent metal with a molar mass of 137 g / mol is barium.

Answer: barium.