Răspunsuri

2014-04-18T11:50:18+03:00
V in c.n = 22,4 litri
CH3OH + 3/2 O2 --->CO2 + 2 H2O
nr moli = m/masa molara=>nr moli metanol = 64/32=2 moli metanol
masa molara CH3OH=12+4+16=32 g/mol
1 mol metanol.......1 mol CO2
2 moli metanol...............x moli CO2          x = 2 moli CO2
V=nr moli * V in cn=2*22,4=>V = 44,8 litri metanol