Ethane & Decane Combustion: CO2 And H2O Yields

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Let's dive into the world of chemistry and figure out how much carbon dioxide (CO2) and water (H2O) are produced when we burn 3 grams of ethane (C2H6) and 5 grams of decane (C10H22). This involves stoichiometry, which is basically the math behind chemical reactions. We'll need to balance the chemical equations, calculate molar masses, and use the mole ratios to find our answers. So, grab your calculators, guys, and let’s get started!

Balancing the Chemical Equations

First, we need to write and balance the combustion equations for both ethane and decane. Combustion involves reacting with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).

Ethane Combustion

The unbalanced equation for ethane combustion is:

C2H6 + O2 → CO2 + H2O

To balance it, we need to make sure the number of atoms for each element is the same on both sides of the equation. The balanced equation is:

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Decane Combustion

The unbalanced equation for decane combustion is:

C10H22 + O2 → CO2 + H2O

Balancing this equation gives us:

2 C10H22 + 31 O2 → 20 CO2 + 22 H2O

Calculating Molar Masses

Next, we need to calculate the molar masses of ethane, decane, carbon dioxide, and water. Molar mass is the mass of one mole of a substance and is usually given in grams per mole (g/mol). We can find these values using the periodic table.

  • Ethane (C2H6): (2 * 12.01) + (6 * 1.01) = 30.08 g/mol
  • Decane (C10H22): (10 * 12.01) + (22 * 1.01) = 142.28 g/mol
  • Carbon Dioxide (CO2): (1 * 12.01) + (2 * 16.00) = 44.01 g/mol
  • Water (H2O): (2 * 1.01) + (1 * 16.00) = 18.02 g/mol

These molar masses are crucial for converting grams to moles and vice versa.

Stoichiometric Calculations for Ethane

Now, let's calculate how much CO2 and H2O are produced from 3 grams of ethane.

Convert Grams of Ethane to Moles

Moles of Ethane = Mass of Ethane / Molar Mass of Ethane Moles of Ethane = 3 g / 30.08 g/mol ≈ 0.0997 mol

Use Mole Ratios from the Balanced Equation

From the balanced equation (2 C2H6 + 7 O2 → 4 CO2 + 6 H2O), we can determine the mole ratios:

  • 2 moles of Ethane produce 4 moles of CO2
  • 2 moles of Ethane produce 6 moles of H2O

So, the mole ratios are:

  • CO2 : Ethane = 4/2 = 2
  • H2O : Ethane = 6/2 = 3

Calculate Moles of CO2 and H2O Produced

Moles of CO2 = Moles of Ethane * Mole Ratio (CO2 : Ethane) Moles of CO2 = 0.0997 mol * 2 ≈ 0.1994 mol

Moles of H2O = Moles of Ethane * Mole Ratio (H2O : Ethane) Moles of H2O = 0.0997 mol * 3 ≈ 0.2991 mol

Convert Moles of CO2 and H2O to Grams

Mass of CO2 = Moles of CO2 * Molar Mass of CO2 Mass of CO2 = 0.1994 mol * 44.01 g/mol ≈ 8.7756 g

Mass of H2O = Moles of H2O * Molar Mass of H2O Mass of H2O = 0.2991 mol * 18.02 g/mol ≈ 5.390 g

Therefore, burning 3 grams of ethane produces approximately 8.7756 grams of CO2 and 5.390 grams of H2O.

Stoichiometric Calculations for Decane

Now, let's do the same calculations for 5 grams of decane.

Convert Grams of Decane to Moles

Moles of Decane = Mass of Decane / Molar Mass of Decane Moles of Decane = 5 g / 142.28 g/mol ≈ 0.0351 mol

Use Mole Ratios from the Balanced Equation

From the balanced equation (2 C10H22 + 31 O2 → 20 CO2 + 22 H2O), we can determine the mole ratios:

  • 2 moles of Decane produce 20 moles of CO2
  • 2 moles of Decane produce 22 moles of H2O

So, the mole ratios are:

  • CO2 : Decane = 20/2 = 10
  • H2O : Decane = 22/2 = 11

Calculate Moles of CO2 and H2O Produced

Moles of CO2 = Moles of Decane * Mole Ratio (CO2 : Decane) Moles of CO2 = 0.0351 mol * 10 ≈ 0.351 mol

Moles of H2O = Moles of Decane * Mole Ratio (H2O : Decane) Moles of H2O = 0.0351 mol * 11 ≈ 0.3861 mol

Convert Moles of CO2 and H2O to Grams

Mass of CO2 = Moles of CO2 * Molar Mass of CO2 Mass of CO2 = 0.351 mol * 44.01 g/mol ≈ 15.4475 g

Mass of H2O = Moles of H2O * Molar Mass of H2O Mass of H2O = 0.3861 mol * 18.02 g/mol ≈ 6.9575 g

Therefore, burning 5 grams of decane produces approximately 15.4475 grams of CO2 and 6.9575 grams of H2O.

Total Amounts of CO2 and H2O Produced

Finally, let's add the amounts of CO2 and H2O produced from both ethane and decane.

Total CO2

Total Mass of CO2 = Mass of CO2 from Ethane + Mass of CO2 from Decane Total Mass of CO2 = 8.7756 g + 15.4475 g ≈ 24.2231 g

Total H2O

Total Mass of H2O = Mass of H2O from Ethane + Mass of H2O from Decane Total Mass of H2O = 5.390 g + 6.9575 g ≈ 12.3475 g

Conclusion

So, when you burn 3 grams of ethane and 5 grams of decane, you get approximately 24.2231 grams of carbon dioxide and 12.3475 grams of water. Stoichiometry can be a bit complex, but breaking it down step-by-step makes it manageable. Always remember to balance your equations, calculate molar masses accurately, and use the correct mole ratios. Keep practicing, and you’ll become a stoichiometry superstar in no time! Remember these are theoretical yields, and actual experimental yields may vary. Also, ensure you are performing such experiments with appropriate safety measures and in a controlled environment. These calculations are fundamental in chemistry for understanding reaction yields and material balances.