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16 January, 13:29

A reaction container holds 5.77 g of P4 and 5.77 g of O2.

The following reaction occurs: P4 + O2 → P4O6.

If enough oxygen is available then the P4O6 reacts further: P4O6 + O2 → P4O10.

a. What is the limiting reagent for the formation of P4O10?

b. What mass of P4O10 is produced?

c. What mass of excess reactant is left in the reaction container?

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Answers (1)
  1. 16 January, 13:52
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    a) O2 is the limiting reactant

    b) 5.75 grams P4O10

    c) 5.79 grams P4O6

    Explanation:

    Step 1: Data given

    Mass of P4 = 5.77 grams

    Mass of O2 = 5.77 grams

    Molar mass of P4 = 123.90 g/mol

    Molar mass O2 = 32.0 g/mol

    Step 2: The balanced equation

    P4 + 3O2 → P4O6

    Step 3: Calculate moles of P4

    Moles P4 = mass P4 / molar mass P4

    Moles P4 = 5.77 grams / 123.90 g/mol

    Moles P4 = 0.0466 moles

    Step 4: Calculate moles O2

    Moles O2 = mass O2 / molar mass O2

    Moles O2 = 5.77 grams / 32.0 g/mol

    Moles O2 = 0.1803 moles

    Step 5: Calculate limiting reactant

    P4 is the limiting reactant in this reaction. It will completely be consumed (0.0466 moles). O2 is in excess, there will react 3*0.0466 = 0.1398 moles

    There will remain 0.1803 - 0.1398 = 0.0405 moles O2

    Step 6: Calculate the amount of P4O6

    For 1 mol P4 we'll have 1 mol P4O6

    For 0.0466 moles P4 we'll have 0.0466 moles P4O6

    Step 7: The balanced equatio

    P4O6 + 2O2 → P4O10

    We have 0.0466 moles P4O6 and 0.0405 moles O2

    Step 8: Calculate the limiting reactant

    For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10

    O2 is the limiting reactant. It will completely be consumed (0.0405 moles)

    P4O6 is in excess. There will react 0.0405/2 = 0.02025 moles

    There will remain 0.0466 - 0.02025 = 0.02635 moles P4O6

    This is 0.02635 * 219.88 g/mol = 5.79 grams P4O6

    Step 9: Calculate moles and mass of P4O10

    For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10

    For 0.0405 moles O2 we'll have 0.02025 moles P4O10

    This is 0.02025 * 283.89 g/mol = 5.75 grams P4O10
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