Question

какой объем водорода выделяется при взаимодействии 16,2 г алюминия с соляной кислотой. помогите пожалуйста))​

Answer 1

2Al + 6HCl = 2AlCl3 + 3H2

n(Al) = 0,6 моль

n(H2) = 0,9 моль

v(H2) = 0,9*22,4=20,16л

Answer 2

Hydrogen Volume Produced from Reaction of Aluminum with Hydrochloric Acid

When 16.2 grams of aluminum reacts with hydrochloric acid, the amount of hydrogen gas produced can be calculated using the stoichiometry of the reaction.

The balanced chemical equation for the reaction between aluminum and hydrochloric acid is:

2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g)

From the balanced equation, it can be seen that 2 moles of aluminum react with 6 moles of hydrochloric acid to produce 3 moles of hydrogen gas.

To find the volume of hydrogen gas produced, we can use the ideal gas law:

PV = nRT

Where: - P = pressure - V = volume - n = number of moles - R = ideal gas constant - T = temperature

Given that the reaction takes place at standard temperature and pressure (STP), which is 0°C (273.15 K) and 1 atm, we can use the molar volume of a gas at STP, which is 22.4 L/mol.

Using the stoichiometry of the reaction, we can calculate the number of moles of hydrogen gas produced from 16.2 grams of aluminum.

The molar mass of aluminum is approximately 27 g/mol. Therefore, 16.2 grams of aluminum is equal to:

16.2 g / 27 g/mol = 0.6 moles of aluminum

According to the stoichiometry of the reaction, 2 moles of aluminum produce 3 moles of hydrogen gas. Therefore, 0.6 moles of aluminum will produce:

(0.6 moles Al) x (3 moles H2 / 2 moles Al) = 0.9 moles of H2

At STP, 1 mole of any gas occupies 22.4 liters. Therefore, 0.9 moles of hydrogen gas will occupy:

0.9 moles x 22.4 L/mole = 20.16 liters of hydrogen gas

So, the volume of hydrogen gas produced from the reaction of 16.2 grams of aluminum with hydrochloric acid is approximately 20.16 liters.