Question

Смесь меди и цинка массой 20 г обработали избытком разбавленной серной кислоты .В результате был собран газ объемом 5,6 дм3. Рассчитайте массовые доли металлов в исходной смеси.

Answer 1

Медь с разбавленной серной не реагирует 
Zn + H2SO4 = H2 + ZnSO4
5.6 дм^3 = 5.6 литров 
ню цинка = 5.6 литров / 22.4 литров на моль = 0.25 моль
m (zn) = 0.25 моль * 65 гр на моль = 16.25 гр
Омега (Zn) = 16.25 гр / 20 гр *100% = 81,25%
Омега меди = 100% - 81,25%= 18.75%

Answer 2

Calculation of Mass Fractions of Copper and Zinc in the Initial Mixture

To calculate the mass fractions of copper and zinc in the initial mixture, we need to use the information provided about the reaction between the mixture of copper and zinc and excess diluted sulfuric acid.

From the given information, we know that: - The mass of the copper-zinc mixture is 20 g. - The volume of the collected gas is 5.6 dm³.

To calculate the mass fractions, we need to determine the moles of copper and zinc in the initial mixture and then divide them by the total moles of the mixture.

Let's start by calculating the moles of the collected gas using the ideal gas law:

PV = nRT

Where: P is the pressure of the gas (assumed to be constant) V is the volume of the gas (5.6 dm³) n is the number of moles of the gas (to be determined) R is the ideal gas constant (8.314 J/(mol·K)) T is the temperature of the gas (assumed to be constant)

Since the pressure and temperature are not given, we can assume standard conditions (1 atm and 273.15 K) for simplicity.

Using the ideal gas law equation, we can solve for n:

n = PV / RT

Substituting the given values:

n = (1 atm * 5.6 dm³) / (0.0821 atm·dm³/(mol·K) * 273.15 K)

Calculating this, we find that n is approximately equal to 0.244 moles.

Now, let's determine the moles of copper and zinc in the initial mixture. Since the reaction between copper, zinc, and sulfuric acid produces copper sulfate and zinc sulfate, we can assume that all the copper and zinc in the initial mixture react completely.

The balanced chemical equation for the reaction is:

2Cu + Zn + 2H₂SO₄ → Cu₂SO₄ + ZnSO₄ + 2H₂O

From the equation, we can see that the ratio of moles of copper to moles of zinc is 2:1.

Let's assume the moles of copper in the initial mixture is x. Therefore, the moles of zinc would be x/2.

The total moles of the mixture can be expressed as:

Total moles = moles of copper + moles of zinc

Total moles = x + x/2

Since the total moles of the mixture are equal to the moles of the collected gas (0.244 moles), we can set up the equation:

x + x/2 = 0.244

Simplifying the equation:

3x/2 = 0.244

Solving for x:

x = (2 * 0.244) / 3

Calculating this, we find that x is approximately equal to 0.1627 moles.

Now, we can calculate the mass of copper and zinc in the initial mixture using their respective molar masses.

The molar mass of copper (Cu) is approximately 63.55 g/mol, and the molar mass of zinc (Zn) is approximately 65.38 g/mol.

Mass of copper = moles of copper * molar mass of copper Mass of copper = 0.1627 moles * 63.55 g/mol

Mass of zinc = moles of zinc * molar mass of zinc Mass of zinc = (0.1627 moles / 2) * 65.38 g/mol

Calculating these values, we find that the mass of copper is approximately 10.34 g, and the mass of zinc is approximately 4.23 g.

Finally, we can calculate the mass fractions of copper and zinc in the initial mixture.

Mass fraction of copper = mass of copper / total mass of the mixture Mass fraction of copper = 10.34 g / 20 g

Mass fraction of zinc = mass of zinc / total mass of the mixture Mass fraction of zinc = 4.23 g / 20 g

Calculating these values, we find that the mass fraction of copper is approximately 0.517, and the mass fraction of zinc is approximately 0.212.

Therefore, the mass fractions of copper and zinc in the initial mixture are approximately 51.7% and 21.2%, respectively.