15 (97.8-2.20)kPa×4.16L nN2)= 8.314kPaL.Kmolx292K =0.164mol NHNO2(s)>2H2O(g)+N2(g) 64.04g 1mol m(NHNO2)=? 0.164mol 64.04g×0.164mo1 m(NHNO2)= 1mol =10.5g 返回主目录 返回次目录 返回主目录 返回次目录 15 NH4NO2 (s) ⎯→ 2H2O(g) + N2 (g) 64.04g 1mol m(NH4NO2 )=? 0.164mol 8.314kPa L K mol 292K (97.8 2.20)kPa 4.16L -1 -1     −  n(N2 ) = 1mol 64.04g0.164mol m(NH4NO2 ) = =10.5g =0.164mol