For the complete transformation of alkali metal weighing 7.8 g into chloride

For the complete transformation of alkali metal weighing 7.8 g into chloride, chlorine with a volume of 224 cm3 was needed. Determine the alkali metal.

1. Let’s compose the reaction equation reflecting the process of chlorination of alkali metals:

2Me + Cl2 = 2MeCl.

2. Let’s calculate the chemical amount of consumed chlorine and, using the found result, determine the amount of alkali metal:

n (Cl2) = V (Cl2): Vm = 2.24: 22.4 = 0.1 mol;

n (Me) = n (Cl2) * 2 = 0.1 * 2 = 0.2 mol.

3. Let’s calculate the relative atomic mass of the desired metal and determine it according to the table of D. I. Mendeleev:

Ar (Me) = m (Me): n (Me) = 7.8: 0.2 = 39;

Me = K.

Answer: K.