Question

Помогите пожалуйста! Объясните как решать?!? При полном сжигании смеси пропана и бутана образовалось 110 г СО2 и 58,5 г Н2О. Сколько процентов пропана содержит смесь? Заранее Спасибо

Answer 1

                        Х         а
С3Н8+5О2= СО2 +Н2О
                       132    72

                              у           в
2С4Н10+13О2= 8СО2 +10Н2О
                             352       180


а=72х/132=0,54х
в= 180у/352=0,51у

х+у=110
а+в=58,5

х+у=110
0,54х+0,51у=58,5
х=75,3(СО2)

у=110-75,3=34,7(масса углекислого газа во 2 уравнении)

 х                   75,3
С3Н8+5О2= СО2 +Н2О
  44               132   

х=25,1
     
    х                       34,2
2С4Н10+13О2= 8СО2 +10Н2О
      116                  352     
у=11,43

масса смеси газа 25,1+11.4=36,5
w(ПРОПАНА)-25,1/36,5=68,78%

Answer 2

Problem Analysis

To solve this problem, we need to determine the percentage of propane in the mixture based on the given information about the products formed during the complete combustion of the mixture.

Given Information

- Amount of carbon dioxide (CO2) formed: 110 g - Amount of water (H2O) formed: 58.5 g

Solution Steps

1. Calculate the molar mass of carbon dioxide (CO2) and water (H2O). 2. Determine the number of moles of carbon dioxide (CO2) and water (H2O) formed. 3. Use the stoichiometry of the balanced chemical equation to relate the moles of carbon dioxide (CO2) and water (H2O) to the moles of propane (C3H8) and butane (C4H10) in the mixture. 4. Calculate the total number of moles of propane (C3H8) and butane (C4H10) in the mixture. 5. Calculate the percentage of propane (C3H8) in the mixture.

Let's proceed with the calculations.

Calculation

1. Molar mass of carbon dioxide (CO2): - Carbon (C) molar mass: 12.01 g/mol - Oxygen (O) molar mass: 16.00 g/mol - Molar mass of CO2 = (12.01 g/mol) + (2 * 16.00 g/mol) = 44.01 g/mol

2. Molar mass of water (H2O): - Hydrogen (H) molar mass: 1.01 g/mol - Oxygen (O) molar mass: 16.00 g/mol - Molar mass of H2O = (2 * 1.01 g/mol) + (16.00 g/mol) = 18.02 g/mol

3. Number of moles of carbon dioxide (CO2) formed: - Moles of CO2 = Mass of CO2 / Molar mass of CO2 = 110 g / 44.01 g/mol

4. Number of moles of water (H2O) formed: - Moles of H2O = Mass of H2O / Molar mass of H2O = 58.5 g / 18.02 g/mol

5. Balanced chemical equation for the complete combustion of propane (C3H8): - C3H8 + 5O2 -> 3CO2 + 4H2O

According to the stoichiometry of the balanced equation, for every 3 moles of carbon dioxide (CO2) formed, there are 1 mole of propane (C3H8) in the mixture.

6. Moles of propane (C3H8) in the mixture: - Moles of C3H8 = (Moles of CO2 / 3)

7. Moles of butane (C4H10) in the mixture: - Moles of C4H10 = (Moles of CO2 / 3) * (4 / 3)

8. Total moles of propane (C3H8) and butane (C4H10) in the mixture: - Total moles = Moles of C3H8 + Moles of C4H10

9. Percentage of propane (C3H8) in the mixture: - Percentage of propane = (Moles of C3H8 / Total moles) * 100

Let's calculate the values.

Calculation Results

- Molar mass of CO2 = 44.01 g/mol - Molar mass of H2O = 18.02 g/mol - Moles of CO2 = 110 g / 44.01 g/mol - Moles of H2O = 58.5 g / 18.02 g/mol - Moles of C3H8 = (Moles of CO2 / 3) - Moles of C4H10 = (Moles of CO2 / 3) * (4 / 3) - Total moles = Moles of C3H8 + Moles of C4H10 - Percentage of propane = (Moles of C3H8 / Total moles) * 100

Now, let's substitute the values and calculate the percentage of propane in the mixture.

Calculation

- Molar mass of CO2 = 44.01 g/mol - Molar mass of H2O = 18.02 g/mol - Moles of CO2 = 110 g / 44.01 g/mol = 2.5 mol - Moles of H2O = 58.5 g / 18.02 g/mol = 3.25 mol - Moles of C3H8 = (2.5 mol / 3) ≈ 0.833 mol - Moles of C4H10 = (2.5 mol / 3) * (4 / 3) ≈ 1.111 mol - Total moles = 0.833 mol + 1.111 mol ≈ 1.944 mol - Percentage of propane = (0.833 mol / 1.944 mol) * 100 ≈ 42.9%

Answer

The mixture contains approximately 42.9% propane.

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