a)
22,4l.............1,5*22,4l..1mol
CH4 + NH3 + 3/2O2 -> HCN + 3H2O
x=?..................y=?........4moli
x=89,6l CH4 pur
p/100=Vpur/Vimpur => Vimpur=89,6*100/98=91,4285l CH4 impur
b)
y=6*22,4=134,4l O2
16g.....44,8l
CH4 + 2O2 -> CO2 + 2H2O
z=?.....134,4l
z=134,4*16/44,8=48g metan
