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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:

  • 2Câ‚‚H₆ + 7Oâ‚‚ → 4COâ‚‚ + 6Hâ‚‚O

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 Oâ‚‚ ÷ 32 g/mol = 3.81 mol Oâ‚‚

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

We calculate the produced moles of COâ‚‚ using the moles of the limiting reactant:

  • 0.83 mol Câ‚‚H₆ * [tex]\frac{4molCO_2}{2molC_2H_6}[/tex] = 1.67 mol COâ‚‚

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

  • 1.67 mol COâ‚‚ * 44 g/mol = 73.48 g

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