What is the percentage yield of N2O5 when 68.0g of O2 reacts With N2 according to the reaction equation, with 72.6g of N2O5 being obtained?
2N2 + 5O2 -> 2N2O5

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  1. I assume that is an excess of N2.
    2N2 + 5O2 -> 2N2O5
    Mols O2 = grams/molar mass = 68.0/32 = approx 2
    mols N2O5 formed if 100% yield = approx 2 x (2/5) = about 0,8.
    grams N2O5 = mols N2O5 x molar mass N2O4 = ? This is the theoretical yield(TY).
    The actual yield (AY) is 72.6 g
    % yield = (AY/TY)*100 = ?
    Check my work. Remember all of the numbers I've shown are estimates and must be recalculated from scratch.

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