Question

Решите задачу! При взаимодействии смеси металлического цинка и цинка карбоната с избытком раствора соляной кислоты выделилось 13.44 л. газа. После полного сжигания образовавшегося газа и конденсации паров объем уменшился до 8.96л. Какова массова доля цинка (%) в исходной смеси?

Answer 1

8,96 л - количество образовавшегося СО2
Разница 13,44-8,96=4,48 л - количество выделившегося водорода
Из уравнения Zn + 2HCl = ZnCl2 + H2 и атомных и молекулярных масс находят кол-во Zn
65,38 * 4,48 / 22,4 = 13,076 г
Из уравнения ZnCO3 + 2HCl = ZmCl2 + H2O + CO2 молекулярных масс находят кол-во ZnCO3
125,38 * 8,96 / 22,4 = 50,125 г ZnCO3

Общая масса смеси Zn+ZnCO3 = 13,076 + 50,125 = 63,201 г дает 100%
Массовая доля цинка в исходной смеси: 13,076*100/63,201=20,69% мас
Массовая доля ZnCO3 в исходной смеси: 50,125*100/63,201=79,31 % 

Answer 2

Problem Analysis

We are given that a mixture of metallic zinc and zinc carbonate reacts with an excess of hydrochloric acid, producing a gas. The volume of gas released is initially 13.44 liters, but after complete combustion and condensation, the volume decreases to 8.96 liters. We need to determine the mass fraction of zinc (%) in the original mixture.

Solution

To solve this problem, we can use the ideal gas law and the stoichiometry of the reaction to find the mass of zinc in the original mixture.

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

Zn + 2HCl -> ZnCl2 + H2

From the equation, we can see that 1 mole of zinc reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.

Let's calculate the number of moles of hydrogen gas released using the ideal gas law:

PV = nRT

Where: P = pressure (assume constant) V = volume of gas (initially 13.44 liters, then 8.96 liters) n = number of moles of gas R = ideal gas constant T = temperature (assume constant)

Since the pressure, gas constant, and temperature are assumed to be constant, we can write:

V1/n1 = V2/n2

Where: V1 = initial volume of gas (13.44 liters) n1 = initial number of moles of gas (to be determined) V2 = final volume of gas (8.96 liters) n2 = final number of moles of gas (1 mole, from the balanced equation)

Solving for n1:

n1 = (V1 * n2) / V2

Now, we can calculate the number of moles of zinc in the original mixture using the stoichiometry of the reaction:

1 mole of zinc = 1 mole of hydrogen gas

Therefore, the number of moles of zinc is equal to the number of moles of hydrogen gas.

Next, we can calculate the molar mass of zinc (Zn) using the periodic table:

Molar mass of Zn = 65.38 g/mol

Finally, we can calculate the mass of zinc in the original mixture using the number of moles of zinc and the molar mass of zinc:

Mass of zinc = number of moles of zinc * molar mass of zinc

To find the mass fraction of zinc (%), we divide the mass of zinc by the total mass of the original mixture and multiply by 100:

Mass fraction of zinc (%) = (mass of zinc / total mass of original mixture) * 100

Let's calculate the mass fraction of zinc now.

Calculation

Given: Initial volume of gas (V1) = 13.44 liters Final volume of gas (V2) = 8.96 liters Molar mass of zinc (Zn) = 65.38 g/mol

Using the ideal gas law, we can calculate the number of moles of hydrogen gas released:

n1 = (V1 * n2) / V2

Substituting the values:

n1 = (13.44 * 1) / 8.96 n1 = 1.5 moles

Since the number of moles of zinc is equal to the number of moles of hydrogen gas, we have:

Number of moles of zinc = 1.5 moles

Now, we can calculate the mass of zinc:

Mass of zinc = number of moles of zinc * molar mass of zinc Mass of zinc = 1.5 moles * 65.38 g/mol Mass of zinc = 98.07 g

To find the mass fraction of zinc (%), we need to determine the total mass of the original mixture. Unfortunately, the problem statement does not provide the total mass of the mixture. Without this information, it is not possible to calculate the mass fraction of zinc.

Please provide the total mass of the original mixture so that we can calculate the mass fraction of zinc.