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Can someone help me with this Determine the volume of oxygen gas produced when 305.80 grams of lead (II) nitrate reacts

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equation: 2Pb(NO3)2 -> 2PbO+4NO2+O2


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For every two moles of Pb(NO3)2, you produce one mole of O2. So, how many moles if 305.80 g? (Hint: Divide by the molecular weight of Pb(NO3)2) Now remember to correct our the stoichiometry (divide by 2). Assuming that you wind up with STP conditions (273K and 1 atmosphere), each mole represents 22.4 L -- so multiply the moles by 22.4 and you answer will be in L.


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