Fe + H2SO4 = FeSO4 + H2
K + H2O = 2KOH + H2
Al + HCI =2AlCI3 + 3H2
Ba + H20 = Ba(OH)2 + H2↑
5CH3-C(CH3)=CH-CH3 + 6KMnO4 + 9H2SO4 ->5CH3-CO-CH3 + 5CH3-COOH + 3K2SO4+ 6MnSO4 + 9H2O
C(0) -2e -> C(+2)
C(-1) - 4e -> C(+3) - 6e | 5| ок-ние; в-ль
Mn(+7) +5e -> Mn(+2) + 5e | 6 | в-ние; ок-ль
CnH2n-2+2HBr->CnH2nBr2
CnH2n-2+2HCl->CnH2nCl2
n(CnH2nBr2)=n(CnH2nCl2)
18,8\(14*n+160)=9,9\(14*n+71)
18,8*(14*n+71)=9,9*(14*n+160)
263,2*n+1334,8=138,6*n+1584
124,6*n=249,2
n=2
C2H2 -ацетилен
2 Ca+O2=2 CaO
5 O2+4P=2 P2O5
O2+4 Li=2 Li2O
2CH3OH + 2Na → 2CH3ONa + H2↑
2C2H5OH + 2Na → 2C2H5ONa + H2↑
m1 + m2 = 11
m1/2*32 + m2/2*46 = 3,36/22,4
m1 + m2 = 11
m1/64 + m2/92 = 0,15
m1 + m2 = 11
92m1 + 64m2 = 883,2
m1 = 11 - m2
92*(11 - m2) + 64m2 = 883,2
1012 - 92m2 + 64m2 = 883,2
28m2 = 128,8
m2 = 4,6 г
m1 = 11 - 4,6 = 6,4 г
w = m1/m
w(CH3OH) = 6,4/11 = 0,582 (58,2%)
w(C2H5OH) = 4,6/11 = 0,418 (41,8%)
Ответ: 58,2% и 41,8%