An alkali metal weighing 1.56 g was placed in a vessel with gaseous chlorine. After the end

An alkali metal weighing 1.56 g was placed in a vessel with gaseous chlorine. After the end of the reaction, the resulting substance was dissolved in water, and a solution of silver nitrate was added. This formed a precipitate weighing 5.74 g. What metal was used for the reaction?

Given:
m (Me) = 1.56 g
m (sediment) = 5.74 g

Find:
Me -?

Solution:
1) Write the equations of chemical reactions:
2Me + Cl2 => 2MeCl;
MeCl + AgNO3 => MeNO3 + AgCl ↓;
2) Calculate the amount of sediment substance – AgCl:
n (AgCl) = m (AgCl) / M (AgCl) = 5.74 / 143.5 = 0.04 mol;
3) Determine the amount of substance Me:
n (AgCl) = n (MeCl) = 0.04 mol;
n (Me) = n (MeCl) = 0.04 mol;
4) Calculate the molar mass of Me:
M (Me) = m (Me) / n (Me) = 1.56 / 0.04 = 39 g / mol;
5) Define Me:
Ar (Me) = M (Me) = 39 – potassium.

Answer: Unknown alkali metal – potassium.