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30 August, 21:39

If 15.0g C3H6, 10.0g O2, and 5.00g NH3 are sealed in a reactor, what mass of acrylonitrile can be produced if the reaction runs to 100% conversion?

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  1. 30 August, 21:53
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    2 C3H6 + 2 NH3 + 3 O2 - > 2 C3H3N + 6 H2O

    Molar masses

    C3H6: 42.08 g/mol

    NH3: 17.03 g/mol

    O2: 32 g/mol

    CH3H3N: 53.06 g/mol

    H2O: 18 g/mol

    # of moles

    C3H6: 15.0 g / [42.08 g/mol] = 0.356 mol

    NH3: 5.00 g / [17.03 g/mol] = 0.294 mol

    O2: 10.0 g / [32 g/mol] = 0.313 mol

    Now wee need to find the limitant reactant

    The theoretical proportion is 2:2:3, equivalent to 1:1:1.5

    The proportion ot the given gases is 0.356 : 0.294 : 0.313, dividing by 0.294 that is equivalent to: 1.2 : 1 : 1.06, where you see that O2 is below the theoretical proportion and it is the limitant reactant.

    Now, from the equation you have 3 mol of O2 produces 2 mol of acrylonitrile (C3H3N)

    Then you will obtain 0.313 * 2/3 = 0.209 mol acrylonitrile

    The mass is 0.209mol * 53.06 g/mol = 11.07 g acrylonitrile

    Answer: 11.07 g
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