`1)`

`n_{Al}={2,7}/{27}=0,1(mol)`

`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`

`0,1->0,15->0,05->0,15(mol)`

`V_{dd\ H_2SO_4}={0,15}/1=0,15(l)=150(ml)`

`->V=150`

`V'=V_{H_2}=0,15.22,4=3,36(l)`

`C_{M\ X}=C_{M\ Al_2(SO_4)_3}={0,05}/{0,15}=1/3M`

`2)`

`n_{Fe}={2,8}/{56}=0,05(mol)`

`Fe+2HCl->FeCl_2+H_2`

`0,05->0,1->0,05->0,05(mol)`

`V_{dd\ HCl}={0,1}/1=0,1(l)=100(ml)`

`->V=100`

`V_{H_2}=0,05.22,4=1,12(l)`

`C_{M\ FeCl_2}={0,05}/{0,1}=0,5M`

a) \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O

Mol:      0,4                         0,2        0,2

b) \(C\%_{ddHCl}=\dfrac{0,4.36,5.100\%}{100}=14,6\%\)

c) \(m_{CaCl_2}=0,2.101=20,2\left(g\right)\)

a) 2Al + 6HCl --> 2AlCl₃ + 3H₂

b) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\); n_HCl = 0,5.2 = 1 (mol)

Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{1}{6}\) => Al hết, HCl dư

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             0,1->0,3---->0,1---->0,15

=> V_H2 = 0,15.22,4 = 3,36 (l)

c) \(\left\{{}\begin{matrix}C_{M\left(AlCl_3\right)}=\dfrac{0,1}{0,5}=0,2M\\C_{M\left(HCl.dư\right)}=\dfrac{1-0,3}{0,5}=1,4M\end{matrix}\right.\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PTHH :

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

0,2     0,4            0,2        0,2 

\(a,V_{H_2}=0,2.22,4=4,48\left(l\right)\)

\(b,m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(m_{ddHCl}=\dfrac{14,6.100}{10}=146\left(g\right)\)

\(c,m_{MgCl_2}=0,2.95=19\left(g\right)\)

\(m_{ddMgCl_2}=4,8+146-\left(0,2.2\right)=150,4\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{19}{150,4}.100\%\approx12,63\%\)

2. 

\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)

\(2K+2H_2O\rightarrow2KOH+H_2\uparrow\)

0,2                     0,2         0,1 

\(m_{KOH}=0,2.56=11,2\left(g\right)\)

\(m_{ddKOH}=7,8+100-\left(0,1.2\right)=107,6\left(g\right)\)

\(C\%=\dfrac{11,2}{107,6}.100\%\approx10,4\%\)

a: \(n_{Zn}=\dfrac{2.6}{65}=0.04\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

0,04                 0,04

\(m_{ZnCl_2}=0.04\left(65+35.5\cdot2\right)=5.44\left(g\right)\)

b: \(n_{HCl}=2\cdot0.04=0.08\left(mol\right)\)

\(m_{ct\left(HCl\right)}=0.08\cdot36.5=2.92\left(g\right)\)

\(C\%\left(HCl\right)=\dfrac{2.92}{200}=0.0146\)

d: \(n_{H_2}=n_{Zn}=0.04\left(mol\right)\)

V=0,04*22,4=0,896(lít)

\(a)n_{Zn}=\dfrac{2,6}{65}=0,04mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

0,04         0,08          0,04         0,04

\(m_{ZnCl_2}=0,04.136=5,44g\\ b)C_{\%HCl}=\dfrac{0,08.36,5}{200}\cdot100=1,46\%\\ c)C_{M\left(HCl\right)}=\dfrac{0,08}{0,2}=0,4M\\ d)V_{H_2}=0,04.22,4=0,896l\\ e)H_2+CuO\underrightarrow{t^0}Cu+H_2O\\ n_{Cu}=n_{H_2}=0,04mol\\ m_{Cu}=0,04.64=2,56g\)

Theo gt ta có: $n_{Al}=0,1(mol)$

a, $2Al+6HCl\rightarrow 2AlCl_3+3H_2$

b, $\Rightarrow n_{H_2}=0,15(mol)\Rightarrow V_{H_2}=3,36(l)$

c, Ta có: $n_{HCl}=0,3(mol)\Rightarrow m_{HCl}=10,95(g)\Rightarrow \%m_{ddHCl}=219(g)$

d, Bảo toàn khối lượng ta có: $m_{dd}=221,4(g)$

$\Rightarrow \%C_{AlCl_3}=6,02\%$

Bài 1:

\(n_{H_2SO_4}=\dfrac{300.19,6\%}{98}=0,6\left(mol\right);n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Vì:\dfrac{0,1}{2}< \dfrac{0,6}{3}\Rightarrow H_2SO_4dư\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{3n_{Al}}{2}=\dfrac{3.0,1}{2}=0,15\left(mol\right)\\ a,m_{Al_2\left(SO_4\right)_3}=342.0,15=51,3\left(g\right)\\ b,m_{ddsau}=m_{Al}+m_{ddH_2SO_4}-m_{H_2}=2,7+300-\dfrac{3}{2}.0,1.2=302,4\left(g\right)\\ c,C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{51,3}{302,4}.100\%\approx16,964\%\\ n_{H_2SO_4\left(dư\right)}=0,6-\dfrac{3}{2}.0,1=0,45\left(mol\right)\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{0,45.98}{302,4}.100\%\approx14,583\%\)

Bài 2:

\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Mg}=b\left(mol\right)\left(a,b>0\right)\\ Hpt:\left\{{}\begin{matrix}27a+24b=7,8\\1,5a+b=\dfrac{8,96}{22,4}=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ \%m_{Al}=\dfrac{0,2.27}{7,8}.100\%\approx69,231\%\Rightarrow\%m_{Mg}\approx100\%-69,231\%\approx30,769\%\)