\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

\(n_{HCl}=\dfrac{7,3}{36,5}=0,2mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(\dfrac{0,2}{2}\) > \(\dfrac{0,2}{6}\)                                      ( mol )

1/15     0,2           1/15        0,1            ( mol )

\(m_{AlCl_3}=\dfrac{1}{15}.13,5=8,9g\)       

\(m_{H_2}=0,1.2=0,2g\)

\(m_{Al\left(dư\right)}=\left(0,2-\dfrac{1}{15}\right).27=3,6g\)

\(a.PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\) 

\(b.n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{m}{M}=\dfrac{3,65}{\left(1+35,5\right)}=0,1\left(mol\right)\) 

Tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,1}{2}\Rightarrow Zn\) dư và dư \(m_{Zn}=n.M=0,2.65=13\left(g\right)\) 

\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)

                 1          2             1         1

              0,05      0,1          0,05     0,05 

\(m_{ZnCl_2}=n.M=0,05.\left(65+35,5.2\right)=6,8\left(g\right)\\ m_{H_2}=n.M=0,05.2=0,1\left(g\right).\)

a)\(PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\)

b)\(n_{Zn}=\dfrac{13}{65}=0,2\left(m\right)\)

\(n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(m\right)\)

\(PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\)

ta có tỉ lệ:\(\dfrac{0,2}{1}>\dfrac{0,1}{2}=>Zn\) dư

\(PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\)

tỉ lệ       :1        2             1             2

số mol  :0,05   0,1          0,05        0,05

\(m_{ZnCl_2}=0,05.136=6,8\left(g\right)\)

\(m_{H_2}=0,05.2=0,1\left(g\right)\)

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

\(1mol\)    \(2mol\)     \(1mol\)      \(1mol\)

\(0,05mol\)  \(0,1mol\)  \(0,05mol\)   \(0,05mol\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(N_{HCl}=\dfrac{m}{M}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\)

\(\text{Ta thấy Zn dư,HCl phản ứng hết}\)

\(m_{ZnCl_2}=n.M=0,05.136=6,8\left(g\right)\)

\(m_{H_2}=n.M=0,05.2=0,1\left(g\right)\)

2Al+6Hcl->2AlCl3+3H2

1\15-0,2------1\15-----0,1 mol   

n Al=0,2 mol

n HCl=0,2 mol

=>al dư

=>m AlCl3=\(\dfrac{1}{15}133,5\)=8,9g

=>m Aldư=\(\dfrac{2}{15}\).27=3,6g

nAl=0,2(mol)

mHCl=500.10%=50(g) => nHCl=50/36,5=100/73(mol)

PTHH: 2 Al + 6 HCl -> 2 AlCl3 +  3 H2

Vì: 0,2/2 < 100/73:6

=> Al hết, HCl dư, tính theo nAl

a) nH2=3/2. 0,2=0,3(mol) => V(H2,đktc)=0,3.22,4=6,72(l)

b) mHCl(tham gia p.ứ)= 6/2. 0,2 . 36,5= 21,9(g)

c) mddsau= 5,4+500-0,3.2=504,8(g)

mAlCl3=0,2. 133,5= 26,7(g)

mHCl(DƯ)=  50 -21,9=28,1(g)

C%ddAlCl3= (26,7/504,8).100=5,289%

C%ddHCl(dư)= (28,1/504,8).100=5,567%

a. Ta có n Al = 5,4:27= 0,2(mol)

Pthh 2Al + 6HCl---> 2AlCl3+3H2

Có nH2= 0,3 => VH2= 0,3*22,4=6,72 

Có n HCl= 0,6 => m HCl= 0,6*36,5 = 21,9 g

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

Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\n_{HCl}=\dfrac{500\cdot10\%}{36,5}=\dfrac{100}{73}\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{\dfrac{100}{73}}{6}\) \(\Rightarrow\) HCl còn dư, Nhôm p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{AlCl_3}=0,2\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\\n_{HCl\left(dư\right)}=\dfrac{11}{15}\left(mol\right)\\n_{HCl\left(p/ứ\right)}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{HCl\left(p/ứ\right)}=0,6\cdot36,5=21,9\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Al}+m_{ddHCl}-m_{H_2}=504,8\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{26,7}{504,8}\cdot100\%\approx5,3\%\\C\%_{HCl\left(dư\right)}=\dfrac{\dfrac{11}{15}\cdot36,5}{504,8}\cdot100\%\approx5,3\%\end{matrix}\right.\) 

$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$

$n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)$

$\Rightarrow n_{Al}=0,15(mol)$

$\Rightarrow \%m_{Al}=\dfrac{0,15.27}{9,45}.100\%\approx 42,86\%$

$\Rightarrow \%m_{Cu}=100-42,86=57,14\%$

$b)$ Theo PT: $n_{HCl}=2n_{H_2}=0,45(mol)$

$\Rightarrow C_{M_{HCl}}=\dfrac{0,45.110\%}{0,5}=0,99M$

\(n_{Fe}=\dfrac{2,8}{56}=0,05mol\)

\(n_{Al}=\dfrac{0,81}{27}=0,03mol\)

\(B:8,12g\left\{{}\begin{matrix}Cu:a\left(mol\right)\\Ag:b\left(mol\right)\\Fe:c\left(mol\right)\end{matrix}\right.\Rightarrow64a+108b+56c=8,12\)(*)

\(n_{H_2}=\dfrac{0,672}{22,4}=0,03mol\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,03                               0,03

\(\Rightarrow c=n_{Fe}=0,03mol\)

Thay vào (*) ta được: \(64a+108b=8,12-0,03\cdot56\left(1\right)\)

\(n_{Fepư}=0,05-0,03=0,02mol\)

\(BTe:2n_{Cu}+n_{Ag}=3n_{Fepư}+2n_{H_2}\)

\(\Rightarrow2a+b=2\cdot0,02+3\cdot0,03\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,05\\b=0,03\end{matrix}\right.\)

\(C_{M_{Cu\left(NO_3\right)_2}}=\dfrac{0,05}{0,1}=0,5M\)

\(C_{M_{AgNO_3}}=\dfrac{0,03}{0,1}=0,3M\)

Đáp án C.