Question

Вычисли массу образовавшейся соли и объем выделявшегося газа если известно что для реакции была взята серная кислота и навеска аллюминия массой 30г.,содержащая 10% примесей. подскжите пожалуйста решение

Answer 1

Сначала найти чистого алюминия: 
ω(Al) = m(Al) / m навески

m(Al) = ω×m навески = 0,1 × 30 =3 г.
                                                         0,11моль              х моль       у моль
Составляем уравнение реакции: 2Al + 3H2SO4 = Al2(SO4)3 + 3H2
                                                       2 моль                   1 моль       3 моль

находим количество вещества Al:  n(Al) = m/M ; n(Al)= 3/27 = 0,11 моль
подставляем в уравнение реакции и находим количество вещества соли и водорода: 
х = 0,11×1 / 2 = 0,055 моль ( Al2(SO4)3)
у = 0,11×3 / 2 = 0,165 моль (Н2)

Находим массу соли:
m(Al2(SO4)3) = n×M(Al2(SO4)3) = 0,055 × 342 = 18,81 г. 
М(Al2(SO4)3) находим по таблице Менделеева = 342 г/моль
Находим объем газа:
∨(Н2) = n × Vm = 0,165 × 22,4 = 3,696 л. ≈3,7 л.

Answer 2

Calculation of Mass of Salt and Volume of Gas Produced

To calculate the mass of the salt and the volume of gas produced in the reaction between sulfuric acid and aluminum, we need to consider the balanced chemical equation for the reaction. However, the specific reaction equation is not provided in the question. Therefore, I will provide a general equation for the reaction and proceed with the calculation based on that.

The balanced chemical equation for the reaction between sulfuric acid (H2SO4) and aluminum (Al) can be represented as follows:

2H2SO4 + 2Al -> Al2(SO4)3 + 3H2↑

From the balanced equation, we can see that 2 moles of sulfuric acid react with 2 moles of aluminum to produce 1 mole of aluminum sulfate (Al2(SO4)3) and 3 moles of hydrogen gas (H2).

Now, let's calculate the mass of the salt and the volume of gas produced based on the given information.

Given: - Mass of aluminum taken = 30 g - Aluminum contains 10% impurities

To calculate the mass of pure aluminum, we need to subtract the mass of impurities from the total mass of aluminum.

Mass of pure aluminum = Total mass of aluminum - Mass of impurities Mass of impurities = 10% of the mass of aluminum

Let's calculate the mass of pure aluminum: Mass of impurities = 10% * 30 g = 3 g Mass of pure aluminum = 30 g - 3 g = 27 g

Now, we can calculate the number of moles of aluminum using its molar mass. The molar mass of aluminum (Al) is approximately 27 g/mol.

Number of moles of aluminum = Mass of pure aluminum / Molar mass of aluminum Number of moles of aluminum = 27 g / 27 g/mol = 1 mol

From the balanced equation, we know that 2 moles of sulfuric acid react with 2 moles of aluminum to produce 1 mole of aluminum sulfate. Therefore, the number of moles of aluminum sulfate produced will also be 1 mole.

Now, let's calculate the mass of aluminum sulfate using its molar mass. The molar mass of aluminum sulfate (Al2(SO4)3) is approximately 342 g/mol.

Mass of aluminum sulfate = Number of moles of aluminum sulfate * Molar mass of aluminum sulfate Mass of aluminum sulfate = 1 mol * 342 g/mol = 342 g

Therefore, the mass of the salt (aluminum sulfate) formed in the reaction is 342 g.

Next, let's calculate the volume of hydrogen gas produced. From the balanced equation, we know that 2 moles of aluminum react to produce 3 moles of hydrogen gas.

Number of moles of hydrogen gas = 3 moles * (Number of moles of aluminum / 2 moles) Number of moles of hydrogen gas = 3 moles * (1 mol / 2 moles) = 1.5 moles

To calculate the volume of gas, we need to use the ideal gas law equation:

PV = nRT

Where: P = Pressure of the gas (assumed to be constant) V = Volume of the gas n = Number of moles of the gas R = Ideal gas constant T = Temperature (assumed to be constant)

Since the pressure and temperature are not provided in the question, we cannot calculate the exact volume of gas produced. However, we can calculate the volume of gas at standard temperature and pressure (STP), which is 0 degrees Celsius and 1 atmosphere.

At STP, 1 mole of any ideal gas occupies approximately 22.4 liters of volume.

Volume of gas produced at STP = Number of moles of hydrogen gas * 22.4 L/mol Volume of gas produced at STP = 1.5 moles * 22.4 L/mol = 33.6 L

Therefore, the volume of gas produced in the reaction is approximately 33.6 liters.

Please note that the above calculations are based on the assumption of a general reaction between sulfuric acid and aluminum. The actual reaction equation may vary depending on the specific conditions and reactants involved.