\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ n_{NaOH}=2.0,2=0,4(mol)\\ Al+NaOH+H_2O\to NaAlO_2+\dfrac{3}{2}H_2(1)\\ Al_2O_3+2NaOH\to 2NaAlO_2+H_2O(2)\\ \Rightarrow n_{Al}=n_{NaOH(1)}=0,1(mol)\Rightarrow m_{Al}=0,1.27=2,7(g)\\ \Rightarrow n_{NaOH(2)}=0,4-0,1=0,3(mol)\\ \Rightarrow n_{Al_2O_3}=0,15(mol)\Rightarrow m_{Al_2O_3}=0,15.102=15,3(g)\\ \Rightarrow \%_{Al}=\dfrac{2,7}{2,7+15,3}.100\%=10\%\\ \Rightarrow \%_{Al_2O_3}=100\%-15\%=85\%\)

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%.

a)

$Fe + 2HCl \to FeCl_2 + H_2$
$FeO + 2HCl \to FeCl_2 + H_2O$

b)

Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$m_{Fe} = 0,15.56 = 8,4(gam)$
$m_{FeO} = 12 - 8,4 = 3,6(gam)$

$n_{FeO} =0,05(mol)$
Theo PTHH : $n_{HCl} = 2n_{Fe} + 2n_{FeO} = 0,4(mol)$

$V_{dd\ HCl} = \dfrac{0,4}{2} = 0,2(lít)$

c) $Fe + CuSO_4 \to FeSO_4 + Cu$

$n_{Cu} = n_{Fe} = 0,15(mol) \Rightarrow m_{chất\ rắn} = m_{FeO} + m_{Cu}$

$= 3,6 + 0,15.64 = 13,2(gam)$

a) PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

b) Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\) \(\Rightarrow n_{Al}=0,1mol\)

\(\Rightarrow m_{Al}=0,1\cdot27=2,7\left(g\right)\) \(\Rightarrow m_{Ag}=7,3\left(g\right)\)

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,1<--0,2<--------------0,1

=> m_Fe = 0,1.56 = 5,6 (g)

=> m_Ag = 32-5,6 = 26,4 (g)

c) \(\left\{{}\begin{matrix}\%Fe=\dfrac{5,6}{32}.100\%=17,5\%\\\%Ag=\dfrac{26,4}{32}100\%=82,5\%\end{matrix}\right.\)

d) \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\)

a) 2Al + 6HCl -> 2AlCl3 + 3H2

Al2O3 + 6HCl -> 2AlCl3 + 3H2O

nH2 = 0,15mol => nAl=0,1mol => mAl=2,7g; mAl2O3 = 10,2g => nAl2O3 = 0,1mol

=>%mAl=20,93% =>%mAl2O3 = 79,07%

b) nHCl = 0,1.3+0,1.6=0,9 mol=>mHCl(dd)=100g

mddY=12,9+100-0,15.2=112,6g

mAlCl3=22,5g=>C%=19,98%

\(\text{Đặt }\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\\ n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ a,PTHH:\left\{{}\begin{matrix}2Al+6HCl\rightarrow2AlCl_3+3H_2\\Fe+2HCl\rightarrow FeCl_2+H_2\end{matrix}\right.\\ b,\text{Theo đề ta có HPT: }\left\{{}\begin{matrix}27x+56y=8,3\\\dfrac{3}{2}x+y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%_{Al}=\dfrac{0,1\cdot27}{8,3}\approx32,53\%\\\%_{Fe}\approx67,47\%\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}n_{AlCl_3}=0,1\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,1\cdot133,5=13,35\left(g\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\end{matrix}\right.\\ \Rightarrow\sum m_{muối}=13,35+12,7=26,05\left(g\right)\)