Question

Помогите пожалуйста!! Сожгли 6,72 л (н.у) пропана (C3H8) и весь образовавшийся при этом углекислый газ поглотили избытком раствора баритовой воды. Найти массу выпавшего осадка.

Answer 1

1.n(смеси) = 6,72/22,4=0,3 моль

2.уравнения.

 1) C2H6 + 3,5 O2 -> 2CO2 + 3H2O

      x моль                 2х моль

  2)C2H8 + 5 O2 -> 3 CO2 + 4 H2О

     у моль               3у моль

3) Са(OH)2 + CO2 -> CaCO3| + H2o

3. считаем общее кол-во СО2

n= m(CaCO3)/M(CaCO3)=80/100=0,8 моль

4. составим систему уравнений

2x + 3y = 0,8

x +y =0,3

следовательно

x = 0,1

y = 0,2

Ответ: 1:2 (этан к пропану)

Answer 2

Calculation of the Mass of the Precipitate

To find the mass of the precipitate formed when 6.72 L (at standard temperature and pressure) of propane (C3H8) is burned and the resulting carbon dioxide gas is absorbed by an excess of barium hydroxide solution, we need to consider the balanced chemical equation for the combustion of propane and the reaction between carbon dioxide and barium hydroxide.

The balanced chemical equation for the combustion of propane is:

C3H8 + 5O2 -> 3CO2 + 4H2O

From the balanced equation, we can see that 1 mole of propane (C3H8) produces 3 moles of carbon dioxide (CO2). We can use this information to calculate the number of moles of carbon dioxide produced.

To calculate the number of moles of carbon dioxide, we need to convert the volume of propane to moles using the ideal gas law. The ideal gas law equation is:

PV = nRT

Where: P = pressure (atm) V = volume (L) n = number of moles R = ideal gas constant (0.0821 L·atm/mol·K) T = temperature (K)

Since the volume of propane is given at standard temperature and pressure (STP), we can use the values: P = 1 atm T = 273 K

Using the ideal gas law equation, we can calculate the number of moles of propane:

n = PV / RT n = (1 atm * 6.72 L) / (0.0821 L·atm/mol·K * 273 K) n ≈ 0.286 moles

Since 1 mole of propane produces 3 moles of carbon dioxide, the number of moles of carbon dioxide produced is:

3 moles CO2 / 1 mole C3H8 * 0.286 moles C3H8 ≈ 0.858 moles CO2

Now, let's consider the reaction between carbon dioxide and barium hydroxide:

CO2 + Ba(OH)2 -> BaCO3 + H2O

From the balanced equation, we can see that 1 mole of carbon dioxide reacts with 1 mole of barium hydroxide to produce 1 mole of barium carbonate (BaCO3). Therefore, the number of moles of barium carbonate formed is also 0.858 moles.

To calculate the mass of the precipitate (barium carbonate), we need to know the molar mass of barium carbonate. The molar mass of BaCO3 is:

1 Ba atom: 1 * 137.33 g/mol = 137.33 g/mol 1 C atom: 1 * 12.01 g/mol = 12.01 g/mol 3 O atoms: 3 * 16.00 g/mol = 48.00 g/mol

Total molar mass of BaCO3 = 137.33 g/mol + 12.01 g/mol + 48.00 g/mol = 197.34 g/mol

Now, we can calculate the mass of the precipitate:

mass = number of moles * molar mass mass = 0.858 moles * 197.34 g/mol ≈ 169.47 g

Therefore, the mass of the precipitate (barium carbonate) formed is approximately 169.47 grams.

Please note that this calculation assumes ideal conditions and complete reaction conversion. In practice, there may be some loss or incomplete conversion of reactants, which could affect the actual mass of the precipitate formed.