Consider the following reaction: 2NO(g)+2H2(g)→N2(g)+2H2O(g)?
a) In the first 60.0 s of this reaction, the concentration of NO dropped from 1.60 M to 0.300 M . Calculate the average rate of the reaction in this time interval.
b) If the volume of the reaction vessel in part (a) was 3.80 L , what mass of N2 was formed during the first 60.0 s of the reaction?
I got the answer to part A (0.0108 Ms^-1) but I can't figure out how to even start part b. Any help would be much appreciated. Thanks in advance :)
- hcbiochemLv 71 month agoFavorite Answer
a) Change of NO = 1.60 - 0.30 M / 60.0 s = 0.0217 M/s
Rate = 0.0217 M/s / 2 = 0.0108 M/sb) Moles NO initial = 3.80 L X 1.60 mol/L = 6.08 mol NOMoles NO at end = 3.80 L X 0.300 M = 1.14 mol NOMoles NO consumed = 6.08 - 1.14 = 4.94 mol4.94 mol NO X (1 mol N2 / 2 mol NO) X 28.0 g/mol N2 = 69.2 g N2