Răspunsuri

  • CH4
  • Ambiţios
2014-11-13T10:36:38+02:00
CH4 + H2O ---> CO + 3 H2
nr moli amestec = 302,4/22,4 = 13,5 moli amestec
nr moli metan = 72/16 = 4,5 moli metan
4 moli amestec........1 mol CO......3 moli H2
13,5 moli amestec.......x moli CO........y moli H2
x = 3,375 moli CO
y = 10,125 moli H2 => V H2 = 10,125 * 22,4 = 226,8 l
16 CH4..........3 moli H2
x g CH4............10,125 moli H2
x = 54 g metan
randamentul = masa practica/masa teoretica * 100 = 54/72 * 100 = 75%
5 3 5