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if 8.6L of H2 reacted with 4.3L of O2 at STP, what is the volume of gaseous water collected (assuming none of it condenses)? 2H2(g) + O2(g) => 2H2O(g)

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    When all are gases one can use L as if they were moles.
    8.6L H2 x (2 moles H2O/2 moles H2) = 8.6*1 = 8.6L

    4.3L O2 x (2 moles H2O/1 mole O2) = 4.3*2 = 8.6.
    This is a limiting reagent problem in which NEITHER is limiting; i.e., you will form 8.6 L H2O as a gas.

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