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buffyoumofo buffyoumofo
wrote...
Posts: 126
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12 years ago
the reaction between nitric oxide(NO) and oxygen to form nitrogen dioxide (NO2) is a key step in photochemical smog formation:

2NO(g) + O2(g) ------------ 2NO(g)

-How many grams of O2 are needed to produce 2.21 g of NO2?
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wrote...
12 years ago
2*molecular weight of NO reacts with 16 gms of O2.
Therefore 2.21 gms reacts with16/15*2.21gms.
wrote...
12 years ago
Calculate the mol of NO2 in 2.21g:

2.21 g * (1 mol/46g) = 0.048 mol NO2

Since the ratio od NO2 : O2 is 2:1 there will be half the amount of O2 required to make this amount of NO2.  So mol of O2 needed = 0.048/2 = 0.024 mol of O2

0.024 mol O2 * (32 g/mol) = 0.77 g of O2 are needed
wrote...
12 years ago
The above is only correct if you assume the reaction favors product only. There is an equilibrium that is created.
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