Consider the reaction for the production of NO2 from NO:

1 answer:

4 0

<span> so now we use the number of moles of NO 5.27 moles. use again the PV=nRT R=62.36367(L mmHg K−1 mol−1) T=35C + 273 = 308K P=632 mm Hg V=97.3 L n=632*97.3/62.3636367*308=3.20145770868 Look at the mole ratio of NO to NO2 its 2:2 they share the same number of moles. Atomic weight of each element in NO2 N:14 O:16 Molar Mass:(16*2)+(14*1)=46 g Actual number of moles=3.20 Theortical number of moles=5.20 Percentage Yield=(Actual Yield/Theortical Yield)*100</span>

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