giúp minh với các bạn 

a, Cu không tác dụng với dd HCl.

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)

\(\Rightarrow m_{Cu}=19,4-13=6,4\left(g\right)\)

c, Ta có: \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{19,4}.100\%\approx67,01\%\\\%m_{Cu}\approx32,99\%\end{matrix}\right.\)

Bạn tham khảo nhé!

\(Zn+2HCl->ZnCl_2+H_2\\ Mg+2HCl->MgCl_2+H_2\\ n_{Zn}=a\\ n_{Mg}=b\\ 65a+24b=11,3g\\ n_{H_2}=a+b=\dfrac{6,72}{22,4}=0,3\\ a=0,1\\ m_{Zn}=65.0,1=6,5g\)

\(a.Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ b.n_{H_2}=\dfrac{4,48}{22,4}-0,2mol\\ n_{Zn}=n_{H_2SO_4}=n_{H_2}=0,2mol\\ \%m_{Zn}=\dfrac{0,2.65}{19,4}\cdot100=67,01\%\\ \%m_{Cu}=100-67,01=32,99\%\\ m_{ddH_2SO_4}=\dfrac{0,2.98}{20}\cdot100=98g\)

a) Các PTHH xảy ra:

\(Mg+2HCl\rightarrow MgCl_2+H_2\) 

\(0,1--0,2------0,1\)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

\(0,1---0,2\) 

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

\(\Rightarrow m_{Mg}=0,1\cdot24=2,4\left(g\right)\)  

\(\Rightarrow m_{ZnO}=10,5-2,4=8,1\left(g\right)\) 

c) \(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)

\(\Rightarrow n_{HCl}=0,2+0,2=0,4\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{14,6\cdot100}{20}=73\left(g\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{73}{1,1}\approx66,4\left(cm^3\right)\)

a. PTHH:

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

Cu + H₂SO₄ ---x--->

b. Theo PT: \(n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.\dfrac{6,72}{22,4}=0,2\left(mol\right)\)

=> \(m_{Al}=0,2.27=5,4\left(g\right)\)

=> \(m_{Cu}=10-5,4=4,6\left(g\right)\)

c. \(\%_{m_{Al}}=\dfrac{5,4}{10}.100\%=54\%\)

\(\%_{m_{Cu}}=100\%-54\%=46\%\)

d. Theo PT: \(n_{H_2SO_4}=n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{29,4}{m_{dd_{H_2SO_4}}}.100\%=20\%\)

=> \(m_{dd_{H_2SO_4}}=147\left(g\right)\)

a) PTHH : \(Fe+2HCl-t^o->FeCl_2+H_2\)  (1)

                 \(2Fe+3Cl_2-t^o->2FeCl_3\)          (2)

                  \(Cu+Cl_2-t^o->CuCl_2\)              (3)

b) Theo pthh (1) : \(n_{Fe}=n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)

Theo pthh (2) và (3) : \(\Sigma n_{Cl2}=\dfrac{3}{2}n_{Fe}+n_{Cu}\)

\(\Rightarrow\dfrac{6,72}{22,4}=\dfrac{3}{2}.0,1+n_{Cu}\)

\(\Rightarrow0,3=0,15+n_{Cu}\)

\(\Rightarrow n_{Cu}=0,15\left(mol\right)\)

\(\Rightarrow m_{Cu}=0,15.64=9,6\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{5,6+9,6}\cdot100\%\approx36,84\%\\\%m_{Cu}=100\%-36,84\%=63,16\%\end{matrix}\right.\)

a) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)  (1)

                \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)  (2)

b) Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\\Sigma n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{Zn}=n_{ZnO}=n_{H_2SO_4\left(1\right)}=n_{H_2SO_4\left(2\right)}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,1\cdot65=6,5\left(g\right)\\m_{ZnO}=0,1\cdot81=8,1\left(g\right)\end{matrix}\right.\)

c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Zn}=n_{ZnSO_4\left(1\right)}=0,1mol\\n_{ZnO}=n_{ZnSO_4\left(2\right)}=0,1mol\end{matrix}\right.\)

\(\Rightarrow\Sigma n_{ZnSO_4}=0,2mol\) \(\Rightarrow C_{M_{ZnSO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

(Coi như thể tích dd thay đổi không đáng kể)

\(a)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=0,1mol\\ m_{Zn}=0,1.65=6,5g\\ m_{Cu}=9,7-6,5=3,2g\\ b)C_{\%ZnCl_2}=\dfrac{0,1.136}{6,5+120-0,1.2}\cdot100=10,77\%\)