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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 25. g of ethane is mixed with 122. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.

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Answer:

73.48 g

Explanation:

The reaction that takes place is:

  • 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

We convert the given reactant masses to moles, using their respective molar masses:

  • 25 g ethane ÷ 30 g/mol = 0.83 mol ethane
  • 122 g O₂ ÷ 32 g/mol = 3.81 mol O₂

0.83 moles of ethane would react completely with (0.83*7/2) 2.9 moles of O₂. There are more O₂ moles than that, so O₂ is the reactant in excess and ethane is the limiting reactant.

We calculate the produced moles of CO₂ using the moles of the limiting reactant:

  • 0.83 mol C₂H₆ * [tex]\frac{4molCO_2}{2molC_2H_6}[/tex] = 1.67 mol CO₂

Finally we convert CO₂ moles to grams, using its molar mass:

  • 1.67 mol CO₂ * 44 g/mol = 73.48 g

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