a

PTHH:

\(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)

\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)

b

Gọi \(\left\{{}\begin{matrix}n_{Al\left(OH\right)_3}=x\left(mol\right)\\n_{Fe\left(OH\right)_3}=y\left(mol\right)\end{matrix}\right.\)

Theo PTHH suy ra: \(\left\{{}\begin{matrix}n_{Al_2O_3}=0,5x\left(mol\right)\\n_{Fe_2O_3}=0,5y\left(mol\right)\end{matrix}\right.\)

Theo đề có hệ phương trình: \(\left\{{}\begin{matrix}78x+107y=29,2\\102.0,5x+160.0,5y=21,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al\left(OH\right)_3}=\dfrac{78.0,1.100\%\%}{29,2}=26,71\%\\\%m_{Fe\left(OH\right)_3}=\dfrac{107.0,2.100\%}{29,2}=73,29\%\end{matrix}\right.\)

a)

$Cu(OH)_2 \xrightarrow{t^o} CuO + H_2O$
$2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O$

Vì theo bảo toàn khối lượng : 

$m_{hh} = m_{oxit} + m_{H_2O}$

mà $m_{H_2O} > 0$ nên $m_{hh} > m_{oxit}$

Do đó khối lượng rắn giảm.

b)

Gọi $n_{Cu(OH)_2} = a ; n_{Fe(OH)_3} = b$

$\Rightarrow 98a + 107b = 6,06(1)$

Theo PTHH : 

$m_{cr} = 80a + 80b = 4,8(2)$
Từ (1)(2) suy ra a = 0,04;   b = 0,02

Suy ra : 

$\%m_{Cu(OH)_2} = \dfrac{0,04.98}{6,06}.100\% = 64,67\%$

$\%m_{Fe(OH)_3} = 35,33\%$

\(n_{Cu\left(OH\right)_2}=a\left(mol\right),n_{Fe\left(OH\right)_3}=b\left(mol\right)\)

\(m_{hh}=98a+107b=6.06\left(g\right)\left(1\right)\)

\(Cu\left(OH\right)_2\underrightarrow{^{^{t^0}}}CuO+H_2O\)

\(a............a\)

\(2Fe\left(OH\right)_3\underrightarrow{^{^{t^0}}}Fe_2O_3+3H_2O\)

\(b............0.5b\)

\(m_{Cr}=80a+0.5b\cdot160=4.8\left(g\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.04,b=0.02\)

\(\%Cu\left(OH\right)_2=\dfrac{0.04\cdot98}{6.06}\cdot100\%=64.68\%\)

\(\%Fe\left(OH\right)_3=35.32\%\)

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,15<-0,3<----0,15<---0,15

\(\left\{{}\begin{matrix}\%Fe=\dfrac{0,15.56}{12}.100\%=70\%\%\\\%Cu=100\%-70\%=30\%\end{matrix}\right.\)

c) m_HCl = 0,3.36,5 = 10,95 (g)

=> \(m_{dd}=\dfrac{10,95.100}{10}=109,5\left(g\right)\)

d) mdd  = 12 + 109,5 - 0,15.2 = 121,2 (g)

 \(C\%\left(FeCl_2\right)=\dfrac{0,15.127}{121,2}.100\%=15,718\%\)

a: \(Cu+2HCl->CuCl_2+H_2\)

\(Fe+2HCl->FeCl_2+H_2\)

Lượng khí hidro thoát ra là 0,15 mol.

a) Al₂O₃ + 2NaOH \(\rightarrow\) 2NaAlO₂ + H₂O

2Al + 2NaOH + 2H₂O \(\rightarrow\) 2NaAlO₂ + 3H₂.

b) Lượng Al phản ứng bằng 2/3 số mol khí hidro thoát ra và bằng 0,1 mol.

\(\Rightarrow\) %m_nhôm=\(\dfrac{27.0,1}{12,9}\).100%\(\approx\)20,93%, %m_{nhôm oxit}\(\approx\)100%-20,93%\(\approx\)79,07%.

Chào em, em đăng bài nhầm môn, em đăng sang môn Hóa nha!

2Fe(OH)3 -----to---> Fe2O3 + 3H2O

Mg(OH)2 ----to---> MgO + H2O

Gọi x, y lần lượt là số mol Fe(OH)3 và Mg(OH)2

\(\left\{{}\begin{matrix}107x+58y=16,5\\\dfrac{1}{2}.160x+y.40=12\end{matrix}\right.\)

=> x=0,1 ; y=0,1

\(\%m_{Fe\left(OH\right)_3}=\dfrac{107.0,1}{16,5}.100=64,85\%\)

%Mg(OH)2 = 35,15%

b) \(2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\)

           0,1----------------------------------->0,05

\(Mg\left(OH\right)_2+H_2SO_4\rightarrow MgSO_4+2H_2O\)

0,1------------------------------------>0,1

\(m_{ddsaupu}=16,5+200=216,5\left(g\right)\)

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,05.400}{216,5}.100=9,24\%\)

\(C\%_{MgSO_4}=\dfrac{0,1.12}{216,5}.100=5,54\%\)

a) PTHH: 2Al + 6 HCl -> 2 AlCl3 + 3 H2

x___________3x______________1,5x(mol)

Fe +2 HCl -> FeCl2 + H2

y___2y____y______y(mol)

b) Ta có: m(rắn)= mCu=0,4(g)

=> m(Al, Fe)=1,5-mCu=1,5-0,4=1,1(g)

nH2= 0,04(mol)

Ta lập hpt:

\(\left\{{}\begin{matrix}27x+56y=1,1\\1,5x+y=0,04\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,02\\y=0,01\end{matrix}\right.\)

=> mAl=27.0,02=0,54(g)

mFe=56.0,01=0,56(g)