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\(\text{Đ}\text{ặt}:n_{Al}=a\left(mol\right);n_{Cu}=b\left(mol\right)\left(a,b>0\right)\\ Al+6HNO_3\rightarrow Al\left(NO_3\right)_3+3NO_2+3H_2O\\ Cu+4HNO_3\rightarrow Cu\left(NO_3\right)_2+2NO_2+2H_2O\\ \Rightarrow\left\{{}\begin{matrix}27a+64b=7,75\\3.22,4a+2.22,4b=7,84\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,1\end{matrix}\right.\\ a,\Rightarrow\%m_{Al}=\dfrac{0,05.27}{7,75}.100\approx17,419\%\\ \Rightarrow\%m_{Cu}\approx82,581\%\\ b,n_{HNO_3}=6a+4b=0,7\left(mol\right)\\ C_{M\text{dd}HNO_3}=\dfrac{0,7}{0,14}=5\left(M\right)\)

a, \(\left\{{}\begin{matrix}27n_{Al}+64n_{Cu}=2,46\\3n_{Al}+2n_{Cu}=3n_{NO}=0,12\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,02\left(mol\right)\\n_{Cu}=0,03\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,54\left(g\right)\\n_{Cu}=1,92\left(g\right)\end{matrix}\right.\)

b, \(n_{HNO_3}=4n_{NO}=0,16\left(mol\right)\)

\(\Rightarrow m_{HNO_3}=10,08\left(g\right)\)

\(\Rightarrow m_{ddHNO_3}=201,6\left(g\right)\)

\(\Rightarrow m_{\text{dd sau pư}}=2,46+201,6-0,04.30=202,86\left(g\right)\)

\(\Rightarrow C\%\left(Al\left(NO_3\right)_3\right)=\dfrac{0,02.213}{202,86}.100\%=2,1\%\)

\(\Rightarrow C\%\left(Cu\left(NO_3\right)_2\right)=\dfrac{0,03.188}{202,86}.100\%=2,78\%\)

$n_{NO_2}  = 0,13(mol)$

$n_{Zn} = a(mol) ; n_{Al} = b(mol)$

Ta có :

$m_{hh} = 65a + 27b = 2,11(gam)$
Bảo toàn electron : $2a + 3b = 0,13$

Suy ra a = 0,02 ; b = 0,03

$\%m_{Zn} = \dfrac{0,02.65}{2,11}.100\%  = 61,61\%$

$\%m_{Al} = 100\% -61,61\% = 38,39\%$

a)\(n_{H_2}=\dfrac{19,832}{24,79}=0,8\left(mol\right)\)

\(PTHH:Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\)

tỉ lệ       :1        1               1              1

số mol  :0,8      0,8            0,8          0,8

\(m_{Zn}=0,8.65=52\left(g\right)\)

b)\(m_{H_2SO_4}=0.8.98=78,4\left(g\right)\)

c)\(m_{ZnSO_4}=0,8.161=128\left(g\right)\)

Đáp án C

\(3Ag+4HNO_3\rightarrow3AgNO_3+NO+2H_2O\)

\(Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO+2H_2O\)

\(NO\) là sản phẩm khử duy nhất.

\(\Rightarrow n_{NO}=\dfrac{0,448}{22,4}=0,02mol\)

Ta có: \(\left\{{}\begin{matrix}27n_{Al}+108n_{Ag}=3,51g\\BTe:3n_{Al}+n_{Ag}=3n_{NO}=0,06\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,01mol\\n_{Ag}=0,03mol\end{matrix}\right.\)

\(\%m_{Al}=\dfrac{0,01\cdot27}{3,51}\cdot100\%=7,7\%\)

\(\%m_{Ag}=100\%-7,7\%=92,3\%\)

a, \(Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO+2H_2O\)

\(3Ag+4HNO_3\rightarrow3AgNO_3+NO+2H_2O\)

b, \(\left\{{}\begin{matrix}\underrightarrow{BTE}3n_{Al}+n_{Ag}=\dfrac{0,448.3}{22,4}=0,06\\\underrightarrow{BTKL}27n_{Al}+108n_{Ag}=3,51\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,01\\n_{Ag}=0,03\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%Al=\dfrac{0,01.27}{3,51}.100\%=7,69\%\\\%Ag=\dfrac{0,03.108}{3,51}.100\%=92,31\%\end{matrix}\right.\)

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

\(n_{HCl}=0,2\cdot4=0,8mol\)

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

\(x\)   \(\rightarrow\)   \(3x\)            \(x\)

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

 \(y\)   \(\rightarrow\) \(2y\)            \(y\)

\(\Rightarrow\left\{{}\begin{matrix}27x+65y=11,9\\3x+2y=0,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

a)\(\%m_{Al}=\dfrac{0,2\cdot27}{11,9}\cdot100\%=45,38\%\)

\(\%m_{Zn}=100\%-45,38\%=54,62\%\)

b)\(\Sigma n_{H_2}=\dfrac{3}{2}x+y=\dfrac{3}{2}\cdot0,2+0,1=0,4mol\)

\(V_{H_2}=0,4\cdot22.4=8,96l\)

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