When interacting with water 0.8 g of a metal located in group 2 of the periodic system of chemical

When interacting with water 0.8 g of a metal located in group 2 of the periodic system of chemical elements of D.I.Mendeleev, 448 ml of hydrogen was released (under normal conditions). Name this metal.

1. Let’s write down the reaction of interaction of an alkaline earth metal with water:

Me + 2H2O = Me (OH) 2 + H2 ↑;

2.Calculate the amount of released hydrogen:

n (H2) = V (H2): Vm = 0.448: 22.4 = 0.02 mol;

3. Determine the chemical amount and find the relative atomic mass of the metal:

n (Me) = n (H2) = 0.02 mol;

Ar (Me) = m (Me): n (Me) = 0.8: 0.02 = 40.

Answer: Ca.