Ta có: \(n_{NO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Theo đề bài: \(108n_{Ag}+64n_{Cu}=2,36\)  (1)

Bảo toàn electron: \(n_{Ag}+2n_{Cu}=n_{NO_2}=0,05\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Ag}=0,01\left(mol\right)\\n_{Cu}=0,02\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Ag}=\dfrac{108\cdot0,01}{2,36}\cdot100\%\approx45,76\%\\\%m_{Cu}=54,24\%\end{matrix}\right.\)

Ag + 2HNO3 ⟶ AgNO3 + H2O + NO2

Cu + 4HNO3 → Cu(NO3)2 + 2NO2↑ + 2H2O

\(n_{NO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Gọi x,y lần lượt là số mol Ag, Cu

Ta có : \(\left\{{}\begin{matrix}108x+64y=2,36\\x+2y=0,05\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=0,01\\y=0,02\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Ag}=0,01.108=1,08\left(g\right)\\m_{Cu}=0,02.64=1,28\left(g\right)\end{matrix}\right.\)

=> \(\%m_{Ag}=\dfrac{1,08}{2,36}.100=45,76\%\)

=>%m_Cu = 100 - 45,76 =54,24%

                           Số mol của khí nito đioxxit 

                       n_NO2= \(\dfrac{V_{NO2}}{22,4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

    Pt :          Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O\(|\)

                     1         4                1                 2            2

                   0,025                                       0,05

                               Số mol của đồng 

                         n_Cu = \(\dfrac{0,05.1}{2}=0,025\left(mol\right)\)

                            Khối lượng của đồng

                            m_Cu = n_Cu . M_Cu

                                    = 0,025 . 64

                                    = 1,6 (g)

                              Khối lượng của bạc

                              m_Ag = 2,36 - 1,6

                                      = 0,76 (g)

               ⁰/₀ m_Cu= \(\dfrac{m_{Cu}.100}{m_{hh}}=\dfrac{1,6.100}{2,36}=67,8\)⁰/₀

               ⁰/₀m_Ag = \(\dfrac{m_{Ag}.100}{m_{hh}}=\dfrac{0,76.100}{2,36}=32,2\)⁰/₀

 Chúc bạn học tốt

a) Gọi số mol Cu, Pb là a, b (mol)

=> 64a + 207b = 14,34 (1)

\(n_{NO}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

Cu⁰ - 2e --> Cu⁺²

a-->2a

Pb⁰ - 2e --> Pb⁺²

b-->2b

N⁺⁵ +3e --> N⁺²

      0,18<--0,06

Bảo toàn e: 2a + 2b = 0,18 (2)

(1)(2) => a = 0,03; b = 0,06

=> \(\left\{{}\begin{matrix}\%m_{Cu}==\dfrac{0,03.64}{14,34}.100\%=13,39\%\\\%m_{Pb}=\dfrac{0,06.207}{14,34}.100\%=86,61\%\end{matrix}\right.\)

b) 

\(\left\{{}\begin{matrix}n_{Cu\left(NO_3\right)_2}=0,03\left(mol\right)\\n_{Pb\left(NO_3\right)_2}=0,06\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Cu\left(NO_3\right)_2}=0,03.188=5,64\left(g\right)\\m_{Pb\left(NO_3\right)_2}=0,06.331=19,86\left(g\right)\end{matrix}\right.\)

ai giúp em câu này với ạ

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

\(n_{SO_2}=\dfrac{2,24}{22,4}=0,1mol\)

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

\(\Rightarrow\left\{{}\begin{matrix}24x+64y=4,4\\BTe:2x+2y=2n_{SO_2}=2\cdot0,1=0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)

\(m_{Mg}=0,05\cdot24=1,2g\)

\(m_{Cu}=0,05\cdot64=3,2g\)

\(n_{H_2SO_4}=2n_{Mg}+2n_{Cu}=2\cdot0,05+2\cdot0,05=0,2mol\)

\(m_{H_2SO_4}=0,2\cdot98=19,6g\)

Đặt nMg=x(mol),nCu=y(mol)

Bảo toàn e : 2nMg+2nCu=2nSO2=0,2(mol)

Từ  giải được:

mMg=0,05.24=1,2(g)

mCu=0,05.64=3,2(g)

nH2SO4=2nSO2=0,2(mol)

mH2SO4=0,2.98=19,6(g)

Đặt nMg=x(mol);nCu=y(mol)

→24x+64y=4,4(1)

nSO2=\(\dfrac{2,24}{22,4}\)=0,1(mol)

Bảo toàn e: x+y=nSO2=0,1(2)

Từ (1)(2)→x=y=0,05(mol)

=>mMg=0,05.24=1,2(g)

mCu=0,05.64=3,2(g)

b)

Mg+2H2SO4→MgSO4+SO2+3H2O

Cu+2H2SO4→CuSO4+SO2+2H2O

Tho PT: ∑nH2SO4=2∑nSO2=0,2(mol)

→m_axit=0,2.98=19,6(g)

$n_{NO_2} = \dfrac{3,136}{22,4} = 0,14(mol)$
Gọi $n_{Mg} = a(mol) ; n_{Cu} = b(mol)$
Ta có : 

$m_X = 24a + 64b = 3,28(gam)$
Bảo toàn electron : $2a + 2b = 0,14$

Suy ra a = 0,03 ; b = 0,04

$\%m_{Cu} = \dfrac{0,04.64}{3,28}.100\% = 78,05\%$

\(n_{SO_2}=\dfrac{V_{SO_2}}{22,4}=\dfrac{2,24}{22,4}=0,1mol\)

Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Cu}=y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=24x\\m_{Cu}=64y\end{matrix}\right.\)

\(Mg+2H_2SO_4\rightarrow MgSO_4+SO_2+2H_2O\) 

 x           2x                               x                  ( mol )

\(Cu+2H_2SO_4\rightarrow CuSO_4+SO_2+2H_2O\)

 y            2x                              y                   ( mol )

Ta có:

\(\left\{{}\begin{matrix}24x+64y=4,4\\x+y=0,1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)

\(\Rightarrow m_{Mg}=0,05.24=1,2g\)

\(\Rightarrow m_{Cu}=0,05.64=3,2\)

\(m_{H_2SO_4}=n_{H_2SO_4}.M_{H_2SO_4}=\left(2.0,05+2.0,05\right).98=0,2.98=19,6g\)

\(n_{Al}=a\left(mol\right),n_{Fe}=b\left(mol\right)\)

\(m=27a+56b=6.95\left(g\right)\left(1\right)\)

\(n_{NO_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

Bảo toàn e : 

\(3a+3b=0.45\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.05,b=0.1\)

\(m_{Al}=0.05\cdot27=1.35\left(g\right)\)

\(m_{Fe}=0.1\cdot56=5.6\left(g\right)\)