Question

Вычислите массу соли и объема (н.у.) водорода , образующие при взаимодействии 3 моль алюминия с избытком соляной кислоты

Answer 1

2Al+6HCl=2AlCl3+3H2

На 2 моля алюминия образуется 2 моля хлорида алюминия, значит на 3 моля - 3 моля соли или 3*М(AlCl3)=3*(27+3*35,5)=400,5 г. На 2 моля алюминия образуется 3 моля водорода, тогда на 3 моля образуется 3/2*3=4,5 моля водорода или 4,5*22,4=100,8 л водорода

Answer 2

Calculating the Mass of Salt and Volume of Hydrogen Gas

When aluminum reacts with an excess of hydrochloric acid, it forms aluminum chloride and hydrogen gas. To calculate the mass of salt (aluminum chloride) and the volume of hydrogen gas produced when 3 moles of aluminum react with an excess of hydrochloric acid, we can use the following chemical equation:

2 Al + 6 HCl → 2 AlCl3 + 3 H2

From the balanced chemical equation, we can see that 2 moles of aluminum react with 6 moles of hydrochloric acid to produce 2 moles of aluminum chloride and 3 moles of hydrogen gas.

Calculating the Mass of Salt (Aluminum Chloride)

The molar mass of aluminum chloride (AlCl3) is approximately 133.34 g/mol.

To calculate the mass of aluminum chloride produced, we can use the following steps:

1. Calculate the moles of aluminum chloride produced. 2. Multiply the moles of aluminum chloride by its molar mass to find the mass.

Step 1: Calculate the Moles of Aluminum Chloride Produced The moles of aluminum chloride produced can be calculated using the stoichiometry of the reaction.

3 moles of aluminum will produce (2/2) * 3 = 3 moles of aluminum chloride.

Step 2: Calculate the Mass of Aluminum Chloride The mass of aluminum chloride produced can be calculated by multiplying the moles of aluminum chloride by its molar mass.

Mass = moles * molar mass Mass = 3 moles * 133.34 g/mol = 400.02 g

Therefore, the mass of aluminum chloride produced is approximately 400.02 grams.

Calculating the Volume of Hydrogen Gas

To calculate the volume of hydrogen gas produced, we can use the ideal gas law, which states:

PV = nRT

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

Since the pressure, temperature, and volume are not specified, we can assume standard conditions (STP) for gases, where the pressure is 1 atm and the temperature is 273.15 K.

Using the molar volume of a gas at STP (22.4 L/mol), we can calculate the volume of hydrogen gas produced.

Step 1: Calculate the Moles of Hydrogen Gas Produced From the balanced chemical equation, 3 moles of aluminum produce 3 moles of hydrogen gas.

Step 2: Calculate the Volume of Hydrogen Gas Volume = moles * molar volume at STP Volume = 3 moles * 22.4 L/mol = 67.2 L

Therefore, the volume of hydrogen gas produced is approximately 67.2 liters.