Question

Срочно!!! Смесь цинка и меди массой 40 г обработали соляной кислотой, при этом выделилось 8,96л газа,определите массовые доли компонентов в смеси.

Answer 1

Дано:
m=40г
V(H2)=8.96л
Найти:
w(Cu,Zn)-?
Решение:
Zn+2HCl=ZnCl2+H2(стрелка вверх,т.к.это газ)
Cu+2HCl=CuCl2+H2(стрелка вверх)
n(H2) = V / Vm = 8.96л / 22,4л/моль= 0,4моль
По уравнению реакции:
0,4                                 0,4
Zn         + 2HCl=ZnCl2+   H2
1 моль                            1 моль
m(Zn)=n*M=0.4*65=26г
w(Zn)=m(Zn)/m(всей смеси)=26/40=0,65=65%
Тоже самое делаем и с медью:
0,4                                0,4
Cu      +2HCl=CuCl2+     H2
1 моль                            1 моль
m(Cu)=n*M=0.4*63.5=25.4г
w(Cu)=m(Cu)/m(всей смеси)=25.4/40=0,635=63.5%

Answer 2

Calculation of Mass Fractions of Zinc and Copper in the Mixture

To determine the mass fractions of zinc and copper in the mixture, we can use the ideal gas law and stoichiometry. The balanced chemical equation for the reaction between zinc and hydrochloric acid is:

Zn + 2HCl -> ZnCl2 + H2

From the given information, we know that 8.96 liters of gas were produced during the reaction. To calculate the mass fractions, we need to find the moles of hydrogen gas produced, which is equal to the moles of zinc reacted.

Using the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature, we can rearrange the equation to solve for n:

n = PV / RT

Given that the volume of gas produced is 8.96 liters, we can convert it to moles using the ideal gas constant R = 0.0821 L·atm/(mol·K) and assuming the temperature is constant at room temperature (around 298 K).

Now, let's calculate the moles of zinc reacted:

n(Zn) = (8.96 L) / (0.0821 L·atm/(mol·K) * 298 K)

Next, we need to convert the moles of zinc to mass using the molar mass of zinc (Zn), which is approximately 65.38 g/mol.

m(Zn) = n(Zn) * M(Zn)

Finally, we can calculate the mass fraction of zinc (w(Zn)) and copper (w(Cu)) in the mixture:

w(Zn) = m(Zn) / (m(Zn) + m(Cu)) w(Cu) = m(Cu) / (m(Zn) + m(Cu))

Please note that the molar mass of copper (Cu) is approximately 63.55 g/mol.

Let's perform the calculations:

Given: - Mass of the mixture = 40 g - Volume of gas produced = 8.96 L

Using the ideal gas law, we can calculate the moles of zinc reacted:

n(Zn) = (8.96 L) / (0.0821 L·atm/(mol·K) * 298 K) = 0.387 mol

Next, we can calculate the mass of zinc reacted:

m(Zn) = n(Zn) * M(Zn) = 0.387 mol * 65.38 g/mol = 25.32 g

To find the mass of copper in the mixture, we subtract the mass of zinc from the total mass:

m(Cu) = 40 g - 25.32 g = 14.68 g

Finally, we can calculate the mass fractions of zinc and copper:

w(Zn) = m(Zn) / (m(Zn) + m(Cu)) = 25.32 g / (25.32 g + 14.68 g) ≈ 0.633 or 63.3% w(Cu) = m(Cu) / (m(Zn) + m(Cu)) = 14.68 g / (25.32 g + 14.68 g) ≈ 0.367 or 36.7%

Therefore, the mass fraction of zinc in the mixture is approximately 63.3%, and the mass fraction of copper is approximately 36.7%.

Please note that these calculations assume ideal conditions and may not account for any impurities or side reactions that could affect the results.