Question

Сероводород объёмом 23,2л пропустили через 1 л 18% раствора сульфата меди. Определить массу осадка. Ответ(99,4г)

Answer 1

Решение:

1) Найдем n(CuSO4)(р-ра)

n(CuSO4)=V/Vm=1/22,4=0,0446 моль

2) Найдем m(CuSO4)(р-ра)

m(CuSO4)=n*M=0,0446*160=7 грамм

3) Найдем m(CuSO4)

m(CuSO4)=w*m(р-ра)=0,18*7=1,26 грамм

4) Найдем n(CuSO4)

n(CuSO4)=1,26/160=0,008 моль

5) Найдем n(H2S)

n(H2S)=V/Vm=23,2/22,4=1,0357 моль

6) По ур-ю реакции найдем n(CuS), считаем по n(H2S)

H2S + CuSO4 → CuS↓ + H2SO4

n(CuS)=1,0357 моль

7) Найдем m(CuS)

m(CuS)=n*M=1,0357*96=99,4 грамм

Ответ: m(CuS)=99,4 грамм

Answer 2

H2S+CuSO4=H2SO4+CuS

 

n(H2S)=23.2л/22,4л/моль=1 моль

m(CuSO4)=(1л*18%)/100%=0,18л

n(CuSO4)=0.18л/22,4л/моль=0,01моль

 

решаем задачу по недостатку:

 

n(CuSO4)=n(CuS)

m(CuS)=0.01моль*95г/моль=0,95г

 

Answer 3

Calculation of Mass of Precipitate

To calculate the mass of the precipitate formed when 23.2 liters of hydrogen sulfide gas is passed through 1 liter of an 18% solution of copper sulfate, we need to use the stoichiometry of the reaction between hydrogen sulfide and copper sulfate.

The balanced chemical equation for the reaction is:

H2S + CuSO4 → CuS + H2SO4

From the equation, we can see that 1 mole of hydrogen sulfide reacts with 1 mole of copper sulfate to form 1 mole of copper sulfide.

To calculate the mass of the precipitate, we need to determine the number of moles of copper sulfate in the 1 liter of 18% solution. The concentration of the solution is given as 18%, which means that 18 grams of copper sulfate are present in 100 milliliters of the solution.

Converting the volume of the solution to milliliters: 1 liter = 1000 milliliters

Converting the concentration to grams of copper sulfate: 18% = 18 grams/100 milliliters

Calculating the number of moles of copper sulfate: Number of moles = mass/concentration

Number of moles = (18 grams/100 milliliters) * (1000 milliliters/1 liter) = 180 grams/liter

Now, we can calculate the mass of the precipitate formed using the stoichiometry of the reaction:

Mass of precipitate = number of moles of copper sulfate * molar mass of copper sulfide

The molar mass of copper sulfide (CuS) is the sum of the atomic masses of copper (Cu) and sulfur (S):

Molar mass of CuS = atomic mass of Cu + atomic mass of S

Molar mass of CuS = 63.55 g/mol + 32.07 g/mol = 95.62 g/mol

Now, we can calculate the mass of the precipitate:

Mass of precipitate = (180 grams/liter) * (1 liter) * (1 mol CuS/1 mol CuSO4) * (95.62 g/mol) = 99.4 grams

Therefore, the mass of the precipitate formed is 99.4 grams.