Ответ:
Объяснение:
CH4 -> CH3-OH (частичное окисление)
CH3-OH + CuO -> H-COH + Cu + H2O
H-COH + Ag2O -> HCOOH + 2Ag
HCOOH + HO-CH3 -> HCOOCH3 + H2O
N(k2o)=40\94=0.43 моль
n(koh)=0.86 моль
m(koh)=0.86*56=44.8 г
<span>тут не всё но большинство </span>
<span>CH4+HNO3 = CH3NO2+H2O</span>
<span>CH3NO2+H2=CH3NH2+H2O</span>
<span>4CH3NH2 + 9O2 = 4CO2 + 10H2O + 2N2</span>
<span><span><span>CH3COOH+HCL2 → CH2ClCOOH+HCL(испарение)</span></span></span>
CH2ClCOOH + 2NH3 → NH2
<span> </span><span>N</span><span>H</span>2<span> —</span><span>C</span><span>H</span>2<span>C</span><span>O</span><span>O</span><span>H</span><span> + </span><span>N</span><span>H</span>4<span>Cl</span>