As a result of the combustion of 9.6 g of metals in chlorine, 38 g of a metal chloride of the composition

As a result of the combustion of 9.6 g of metals in chlorine, 38 g of a metal chloride of the composition МеСL2 were obtained. Determine the metal Ме.

Let’s write the reaction equation:

Me + Cl2 = MeCl2;

We are given the mass of an unknown metal and the mass of the reaction product with the same metal. Let us express the quantities of mol (n) Me and MeCl2:

n (Me) = m (Me): M (Me) = 9.6 g: x g / mol (we take the unknown atomic mass of this metal for x).

n (MeCl2) = m (MeCl2): M (MeCl2) = 38 g: (x + 71) g / mol. (71 is the molar mass of Cl2 in MeCl2).

There are no coefficients in the reaction, which means we can equate the amounts of moles of Me and MeCl2:

9.6: x = 38: (x + 71);

We multiply our construction “crosswise” (if we write through a fraction, it will be clearer):

38x = (71 + x) * 9.6;

38x = 681.6 + 9.6x;

28.4x = 681.6;

x = 24.

A 24 is the atomic mass of Mg (according to the periodic table). Thus, Mg is the metal we were looking for.