Question

Який об’єм водню виділиться при взаємодії цинку із хлоридню кислотою масою 19,8 г? С дано срочнооооо

Answer 1

Ответ:

6.07

Объяснение:                    19.8                    x

Дано:                         Zn+ 2HCl ⇒ ZnCl2 + H2

m(HCl)=19.8г                      73                     22.4

___________

V(H2)-?                   x= 19.8*22.4/73 = 6,07 литров

Answer 2

Reaction of Zinc with Hydrochloric Acid

The reaction of zinc with hydrochloric acid can be represented by the following chemical equation:

Zn + 2HCl → ZnCl2 + H2

This equation shows that when zinc reacts with hydrochloric acid, it forms zinc chloride and hydrogen gas.

Volume of Hydrogen Gas Produced

To calculate the volume of hydrogen gas produced when 19.8g of zinc reacts with hydrochloric acid, we can use the ideal gas law. The molar volume of a gas at standard temperature and pressure (STP) is 22.4 liters per mole.

First, we need to find the number of moles of zinc using its molar mass, which is approximately 65.38 g/mol.

Number of moles of zinc = Mass of zinc / Molar mass of zinc Number of moles of zinc = 19.8g / 65.38 g/mol

Now, we can use the stoichiometry of the reaction to find the number of moles of hydrogen gas produced. According to the balanced chemical equation, 1 mole of zinc produces 1 mole of hydrogen gas.

Finally, we can calculate the volume of hydrogen gas produced using the molar volume at STP.

Volume of hydrogen gas = Number of moles of hydrogen gas * Molar volume at STP Volume of hydrogen gas = Number of moles of zinc * Molar volume at STP

Let's calculate the volume of hydrogen gas produced.