Calculate which weighed portions of magnesium and aluminum must be taken so that when they are dissolved

Calculate which weighed portions of magnesium and aluminum must be taken so that when they are dissolved in hydrochloric acid, the volume of hydrogen released is the same at 12.32 liters

To solve this problem, we write down the given: V (H2) = 12.32 liters.

Find: the mass of aluminum and the mass of magnesium.

Solution:

Let us write the equation for the reaction of the interaction of magnesium with hydrochloric acid.

Mg + HCl = MgCl2 + H2

Let’s arrange the coefficients.

Mg + 2HCl = MgCl2 + H2

Above magnesium we write x g, and under magnesium its molar mass, which is 24 g / mol.

Over hydrogen we write 12.32 liters, and under hydrogen a constant molar volume, which is 22.4 l / mol.

Let’s compose and solve the proportion.

x = 24 * 12.32 / 22.4 = 13.2 g.

Let us write the equation for the reaction of the interaction of aluminum with hydrochloric acid.

Al + HCl = AlCl3 + H2

Let’s arrange the coefficients.

2Al + 6HCl = 2AlCl2 + 3H2

Above aluminum we write x g, and under aluminum – 54 g.

We write 12.32 liters over hydrogen, and 67.2 liters under hydrogen.

Let’s compose and solve the proportion.

x = 54 * 12.32 / 67.2 = 9.9 g

Answer: 9.9 g; 13.2g