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a)
1)\(Fe_2O_3+3H_2SO_4->Fe_2\left(SO_4\right)_3+3H_2O\)
2) \(Fe_2\left(SO_4\right)_3+3BaCl_2->3BaSO_4\downarrow+2FeCl_3\)
3) \(FeCl_3+3NaOH->Fe\left(OH\right)_3\downarrow+3NaCl\)
4) \(Fe\left(OH\right)_3+3HCl->FeCl_3+3H_2O\)
b)
1) \(2Al+6HCl->2AlCl_3+3H_2\)
2) \(AlCl_3+3NaOH->3NaCl+Al\left(OH\right)_3\downarrow\)
3) \(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)
4) \(Al_2O_3+2NaOH->2NaAlO_2+H_2O\)
a) tất cả
\(NaOH+HCl->NaCl+H_2O\)
\(Mg\left(OH\right)_2+2HCl->MgCl_2+2H_2O\)
\(Ba\left(OH\right)_2+H_2O->BaCl_2+H_22O\)
b) Mg(OH)2
\(Mg\left(OH\right)_2-^{t^o}>MgO+H_2O\)
c) NaOH , Ba(OH)2
\(2NaOH+CO_2->Na_2CO_3+H_2O\)
\(Ba\left(OH\right)_2+CO_2->BaCO_3+H_2O\)
d) NaOH,Ba(OH)2
làm quỳ tím hóa xanh
Em tham khảo link này nha!
https://hoc24.vn/cau-hoi/hoa-tan-hoan-toan-121-g-hon-hop-cuo-va-zno-can-100-ml-dung-dich-hcl-3ma-tinh-phan-tram-theo-khoi-luong-moi-oxit-trong-hon-hop-ban-dau-b-neu-hoa.183575761941
Câu 3:
nNaHCO3= (500.20%)/84=25/21(mol)
PTHH: NaHCO3 + CH3COOH -> CH3COONa + CO2 + H2O
nCH3COOH=nNaHCO3= 25/21 (mol)
=> mCH3COOH= 25/21 x 60= 500/7 (g)
=> C%ddCH3COOH= [(500/7)/300).100= 23,81%
b) 2 C4H10 + 5 O2 -to,xt-> 4 CH3COOH + 2 H2O
nC4H10= 2/4 . 25/21= 25/42(mol)
=>V=V(C4H10,đktc)=25/42 . 22,4=13,33(l)
Câu 2a em xem SGK
Câu 2b)
PT 1 phải ra C2H5OH chứ nhở?
$2Na + 2C_2H_5OH \to 2C_2H_5ONa + H_2$
$2Na + 2CH_3COOH \to 2CH_3COONa + H_2$
$CH_3COOH + NaOH \to CH_3COONa + H_2O$
$(C_{17}H_{33}COO)_3C_3H_5 + 3NaOH \to 3C_{17}H_{33}COONa + C_3H_5(OH)_3$
$CaO + 2CH_3COOH \to (CH_3COO)_2Ca + H_2O$
$K_2CO_3 + 2CH_3COOH \to 2CH_3COOK + CO_2 + H_2O$