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2C4H10+1302 → 8CO2 + 10H2O Oxygen is the limiting reactant when 20.0 grams of butane burns in 60.0 grams of oxygen. What mass of excess reactant remains at the completion of the reaction? [?] g C4H10 remain

Question

2C4H10+1302 → 8CO2 + 10H2O Oxygen is the limiting reactant when 20.0 grams of butane burns in 60.0 grams of oxygen. What mass of excess reactant remains at the completion of the reaction? [?] g C4H10 remain

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Solution

To solve this problem, we need to follow these steps:

  1. Convert the mass of the reactants to moles. The molar mass of C4H10 (butane) is approximately 58.12 g/mol and the molar mass of O2 is 32.00 g/mol.

    Moles of C4H10 = 20.0 g / 58.12 g/mol = 0.344 moles Moles of O2 = 60.0 g / 32.00 g/mol = 1.875 moles

  2. Determine the stoichiometric ratio from the balanced chemical equation. The ratio of C4H10 to O2 is 2:13.

  3. Determine the limiting reactant. The ratio of moles of C4H10 to O2 is 0.344:1.875, which simplifies to approximately 1:5.45. This is less than the stoichiometric ratio of 2:13, so O2 is the limiting reactant.

  4. Calculate the amount of excess reactant. The stoichiometric ratio tells us that for every 2 moles of C4H10, we need 13 moles of O2. So for 0.344 moles of C4H10, we need 0.344 * (13/2) = 2.242 moles of O2. We started with 1.875 moles of O2, so we have no O2 left and 2.242 - 1.875 = 0.367 moles of C4H10 left.

  5. Convert the moles of excess reactant back to grams.

    Mass of C4H10 = 0.367 moles * 58.12 g/mol = 21.3 g

So, 21.3 g of C4H10 remain.

This problem has been solved

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