What volume of hydrogen is available at the interaction of 13g zinc at 150 g of a 12% solution of acetic acid?

When dissolved in acetic acid, the metallic zinc turns into zinc acetate. Zinc is a bivalent metal. Therefore, with a reaction, it interacts with a double amount of acetic acid. The single chemical amount of zinc acetate reaction is described by the following chemical reaction equation:

Zn + 2Sn3Cone = Zn (CH3SOO) 2 + H2;

Calculate the chemical amount of acetic acid. For this purpose, we divide the weight of the acid on the weight of 1 praying of this acid.

M CH3Cone = 60 grams / mole;

N CH3Cone = 150 x 0.12 / 60 = 0.3 mol;

Determine the molar amount of zinc. To do this, we split its weight for a mass of 1 mole of metal.

N zn = 13/65 = 0.2 mol;

With such a number of acid, 0.3 / 2 = 0.15 mol zinc is presented. At the same time, 0.15 mol of hydrogen will be synthesized. Find its volume. To do this, multiply the amount of substance to the volume of 1 praying gas (component of 22.4 liters).

V h2 = 0.15 x 22.4 = 3.36 liters;