Hexane (C6H14) burns in air (O2) to give CO2 and H2O

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Hexane (C6H14) burns in air (O2) to give CO2 and H2O. a) Write the balanced equation for this reaction. b) If 215.0 g of hexane is mixed with 215.0 g of oxygen, is hexane or oxygen the limiting reactant? Show work. c) What mass of excess reactant remains at the end of the reaction? d) How many grams of carbon dioxide are produced? I already solved the equation for the problem: 2C6H14 + 19O2 -> 12CO2 + 14H2O. I need help with parts B-D.

The Correct Answer and Explanation is:

Let’s work through parts B-D based on the balanced chemical equation you’ve already written:
2C6H14+19O2→12CO2+14H2O2C_6H_{14} + 19O_2 \rightarrow 12CO_2 + 14H_2O2C6​H14​+19O2​→12CO2​+14H2​O

Part B: Identifying the Limiting Reactant

To find out which reactant is limiting, we’ll need to compare the mole ratios of hexane and oxygen based on the amount of each reactant you have. Here are the steps:

  1. Calculate the moles of hexane (C6H14):
    • Molar mass of hexane (C6H14):
      6×12.01+14×1.008=86.18 g/mol6 \times 12.01 + 14 \times 1.008 = 86.18 \, \text{g/mol}6×12.01+14×1.008=86.18g/mol
    • Moles of hexane:
      215.0 g86.18 g/mol=2.49 mol C6H14\frac{215.0 \, \text{g}}{86.18 \, \text{g/mol}} = 2.49 \, \text{mol} \, \text{C6H14}86.18g/mol215.0g​=2.49molC6H14
  2. Calculate the moles of oxygen (O2):
    • Molar mass of oxygen (O2):
      2×16.00=32.00 g/mol2 \times 16.00 = 32.00 \, \text{g/mol}2×16.00=32.00g/mol
    • Moles of oxygen:
      215.0 g32.00 g/mol=6.72 mol O2\frac{215.0 \, \text{g}}{32.00 \, \text{g/mol}} = 6.72 \, \text{mol} \, \text{O2}32.00g/mol215.0g​=6.72molO2
  3. Compare the ratios from the balanced equation:
    • From the equation, the mole ratio of C6H14 to O2 is 2:19.
      That means for every 2 moles of hexane, you need 19 moles of oxygen.
  4. Determine the amount of oxygen needed for 2.49 moles of hexane:
    • Oxygen needed:
      192×2.49=23.71 mol O2\frac{19}{2} \times 2.49 = 23.71 \, \text{mol O2}219​×2.49=23.71mol O2
    Since we only have 6.72 moles of oxygen available, oxygen is the limiting reactant.

Part C: Mass of Excess Reactant Remaining

Now that we know oxygen is the limiting reactant, we can determine how much hexane is left over:

  1. Find the moles of hexane that will react with the available oxygen:
    • From the equation, 19 moles of oxygen react with 2 moles of hexane.
      So, for 6.72 moles of oxygen:
      219×6.72=0.71 mol C6H14\frac{2}{19} \times 6.72 = 0.71 \, \text{mol C6H14}192​×6.72=0.71mol C6H14
  2. Calculate the mass of hexane that reacts:
    • Mass of hexane reacted:
      0.71 mol×86.18 g/mol=61.2 g0.71 \, \text{mol} \times 86.18 \, \text{g/mol} = 61.2 \, \text{g}0.71mol×86.18g/mol=61.2g
  3. Determine the mass of excess hexane:
    • Initial mass of hexane = 215.0 g
    • Mass of hexane reacted = 61.2 g
    • Mass of excess hexane =
      215.0 g−61.2 g=153.8 g215.0 \, \text{g} – 61.2 \, \text{g} = 153.8 \, \text{g}215.0g−61.2g=153.8g

Thus, 153.8 g of hexane remains after the reaction.

Part D: Mass of Carbon Dioxide Produced

To find the mass of carbon dioxide produced, we can use the amount of limiting reactant (oxygen) and the stoichiometry of the reaction:

  1. Calculate the moles of CO2 produced:
    • From the equation, 19 moles of O2 produce 12 moles of CO2.
      So, for 6.72 moles of oxygen:
      1219×6.72=4.25 mol CO2\frac{12}{19} \times 6.72 = 4.25 \, \text{mol CO2}1912​×6.72=4.25mol CO2
  2. Calculate the mass of CO2 produced:
    • Molar mass of CO2 = 44.01 g/mol
    • Mass of CO2 produced:
      4.25 mol×44.01 g/mol=187.0 g4.25 \, \text{mol} \times 44.01 \, \text{g/mol} = 187.0 \, \text{g}4.25mol×44.01g/mol=187.0g

Therefore, 187.0 g of CO2 is produced in the reaction.

Summary of Answers:

  • Limiting reactant: Oxygen (O2)
  • Mass of excess reactant remaining: 153.8 g of hexane
  • Mass of CO2 produced: 187.0 g

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