Рассчитайте объем кислорода(н.у), который необходим для сжигания 3, 1г фосфора с образованием оксида фосфора(5)

ответ во вложении

Объяснение:

1. C2H6 + Br2 = C2H5Br + HBr
C2H5Br + NaOH = C2H5OH + NaBr
C2H5OH + CH3COOH ---(H2SO4)---> CH3COOC2H5 + H2O
CH3COOC2H5 + NaOH = CH3COONa + C2H5OH
CH3COONa + NaOH + t (сплав) = CH4 + Na2CO3  CH4 + t = C2H2
C2H2 + HCl = CH2=CH-Cl    nCH2=CH-Cl = [-CH2-CH(Cl)-]n

2. CH3COOCH3 ---(NaOH/H2O)---> CH3COONa + CH3OH
CH3OH + O2 ---(V2O5/t)---> HCOH + H2O
HCOH + [Ag(NH3)2]OH = Ag + HCOOH + H2O + 2NH3
HCOH ---(H2/Ni)---> CH3OH
CH3OH + C2H5COOH ---(H2SO4)---> C2H5COOCH3 + H2O
C2H5COOCH3 + NaOH (водн) = C2H5COONa + CH3OH
C2H5COONa + NaOH + t (сплав) = C2H6 + Na2CO3

3. C3H7OH ---(Cu/t)---> C2H5CHO + H2
C2H5CHO + Cu(OH)2 = C2H5COOH + Cu2O
C2H5COOH + C2H5OH ---(H2SO4)---> C2H5COOC2H5
C2H5COOC2H5 + NaOH (водн) = C2H5COONa + C2H5OH

73 г                                                                                 Х л

2HCl            +       Na2CO3       =         2NaCl        +      CO2    +     H2O

n = 2 моль                                                                    n = 1 моль

Mr = 36 г/моль                                                            Vm = 22,4 л/моль

m = 72 г                                                                       V = 22,4 л

73 г HCl    -    Х л СО2

72 г HCl    -    22,4 л СО2

V(CO2) = 73 * 22,4 / 72 = 22,7 л