1)CO2+2Mg----2MgO+C
2C+O2----2CO ( недостаток O2)
2CO+ O2----2CO2
CO2+CaO----CaCO3
CaCO3+H20+CO2----Ca(HCO3)2
Ca(HCO3)2---CaCO3+CO2+H2O
2) XH4, где X - элемент
3)б
4)CaCO3+SiO2----CaSiO3+CO2
m(CaCO3)=120-120*0.1=108 кг
N(CaCO3)=108/100=1.08 кмоль
N(CaSiO3)=N(CaCO3)=1.08 кмоль
M(CaSiO3)=116*108=125.28 г
Ответ : 1.08 кмоль , 125.28 кг
1) friedel crafts acylation (CO + HCl+ AlCl3(cat.) ====> benzaldehyd
2) nitration (HNO3/H2SO4(cat.) ====> 3-nitro-m-benzaldehyd <---> 3-nitro-p-benzaldehyd
3)double wolf-kishner (N2H4 + KOH, t*)===> 3-amino-m/pbenzaldehyd (dynamic stability m- and p- form)
4)sandmeyer rxn I) NaNO2 + HCl II) H3O+ + t*-----> m- p- cresol
p-cresol + NaOH----> C7H7ONa (sodium cresol)
p-cresol + PCl5 -----> C7H6Cl + HCl + POCl3 (1-methyl-4-chlorobenzene
Ответ:
N2 : O2 = 36,84% : 63,16%
14x : 16y = 36,84 : 63,16
x:y = 36,84/14 : 63,16/16 = 2,6 : 3,9
x = 2,6 : 2,6 = 1
y = 3,9 : 2,6 = 1,5
1×2=2 N2
1,5×2=3 O2
N2O3
m<span>Fe3O4= 1000*0,72=720кг </span>
<span>n(<span>Fe3O4)720/232=3.1моль</span></span>
<span><span><span>3Fe3O4 +8Al=4Al2O3+9Fe</span></span></span>
<span><span><span>по уравнению реакции пFe=3т(<span>Fe3O4)=3*3,1=9,3моль</span></span></span></span>
<span><span><span><span>m(Fe)9.3*56=520.8г</span></span></span></span>