дано
m(Fe3O4) = 66 g
w() = 12%
η (Fe) = 84%
-------------------------
mпракт(Fe)-?
m(Fe3O4) = 66 - (66*12% / 100%) = 58.08 g
Fe3O4 +4H2-->3Fe+4H2O
M(Fe3O4) = 232 g/mol
n(Fe3O4) = m/M = 58.08 / 232 = 0.25 mol
n(Fe3O4) = 3n(Fe)
n(Fe) = 3*0.25 = 0.75 mol
M(Fe) = 56 g/mol
mтеор(Fe) = n*M = 0.75 *56 = 42 g
mпракт (Fe) = 42*84% / 100% = 35.28 g
ответ 35.28 г
AL4C3 +12H2O ---->3CH4 + 4Al(OH)3
X1
2CH4 ---->C2H2 +3H2
X2
3C2H2 ---->C6H6
X3
C6H6 + Br2 ---->C6H5Br + HBr
X4
Б)бутанол-2,в) 3- метил-бутанол-2
Дано
CxHy
y = 2x - 2
D(O2) = 2.125
Решение
Mr(CxHy) = D(O2) * Mr(O2) = 2.125 * 32 = 68
Mr(CxHy) = xAr(C) + yAr(H) = 12x + y = 12x + 2x - 2 = 14x - 2 = 68
14x - 2 = 68
14x = 70
x = 5
y = 2 * 5 - 2 = 8
C5H8
Ответ: C5H8