Asked by greg
Ammonia is composed of H and N in a ratio of 9.33g N to 2.00g H. If 4.56g N combine completely with H to form ammonia, how many grams of ammonia would be formed? How do you calculate this?
Answers
Answered by
DrBob222
9.33/atomic mass N = 9.33/14 = 0.666 mol N.
2.00/atomic mass H = 2.00/1 = 2 mol H.
ratio is NH3.
4.56g x (1 mol NH3/1 mol N) x (17g NH3/1 mol NH3) x (1 mol N/14 g N) = ? g NH3.
2.00/atomic mass H = 2.00/1 = 2 mol H.
ratio is NH3.
4.56g x (1 mol NH3/1 mol N) x (17g NH3/1 mol NH3) x (1 mol N/14 g N) = ? g NH3.
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