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

a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

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

Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{10,5}.100\%\approx61,9\%\\\%m_{Cu}\approx38,1\%\end{matrix}\right.\)

c, \(n_{H_2SO_4}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ a.Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ b.n_{Zn}=n_{H_2SO_4}=n_{H_2}=0,1mol\\ m_{Zn}=0,1.65=6,5g\\ \%m_{Zn}=\dfrac{6,5}{10,5}\cdot100\%=61,9\%\\ \%m_{Cu}=100\%-61,9=38,1\%\\ c.C_{M_{H_2SO_4}}=\dfrac{0,1}{0,5}=0,2M\)

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

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

b) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Zn}\) \(\Rightarrow n_{ZnO}=\dfrac{20-0,1\cdot65}{81}=\dfrac{1}{6}\left(mol\right)\)

\(\Rightarrow n_{ZnCl_2}=n_{Zn}+n_{ZnO}=\dfrac{4}{15}\left(mol\right)\) 

Mặt khác: \(m_{H_2}=0,1\cdot2=0,2\left(g\right)\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=119,8\left(g\right)\) \(\Rightarrow C\%_{ZnCl_2}=\dfrac{\dfrac{4}{15}\cdot136}{119,8}\cdot100\%\approx30,27\%\)

c) Giả sử khí là SO₂

PTHH: \(Zn+H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}ZnSO_4+SO_2\uparrow+H_2O\)

Theo PTHH: \(n_{SO_2}=n_{Zn}=0,1\left(mol\right)\) \(\Rightarrow V_{SO_2}=0,1\cdot22,4=2,24\left(l\right)\)

\(A/PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(B/n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ n_{Fe}=a;n_{Zn}=b\\ \Rightarrow\left\{{}\begin{matrix}56a+65b=18,6\\a+b=0,3\end{matrix}\right.\\ \Rightarrow a=0,1;b=0,2\\ m_{Fe}=0,1.56=5,6g\\ m_{Zn}=18,6-5,6=13g\)

a) Đồng không phản ứng được với dung dịch HCl

Zn   + 2HCl  →  ZnCl₂  +  H₂

b) nH₂ = \(\dfrac{2,24}{22,4}\)= 0,1 mol = nZn

=> mZn = 0,1.65 = 6,5 gam , mCu = 21 - 6,5 = 14,5 gam

Vì kẽm phản ứng hết với HCl nên chất rắn thu được sau phản ứng chỉ còn đồng có khối lượng là 14,5 gam

C1:

nH2= 0,15(mol)

PTHH: Fe + 2 HCl-> FeCl2 + H2

0,15_____0,3______0,15___0,15(mol)

-> mFe= 0,15.56=8,4(g)

-> mCuO= 10-8,4=1,6(g)

b) -> nCuO= 0,02(mol)

PTHH: CuO +2 HCl -> CuCl2 + H2O

0,02_________0,04(mol)

c) nHCl (tổng)= 0,34(mol)

=> CMddHCl= 0,34/0,2=1,7(M)

C2:

a) Mg +2 HCl -> MgCl2 + H2

nH2= 0,2(mol) -> nMg= nMgCl2= nH2=0,2(mol); nHCl=0,4(mol)

mMg= 0,2.24=4,8(g) -> mCu= 5,2(g)

=> %mMg=(4,8/10).100=48%

=>%mCu= 100%-48%=52%

b) mMgCl2= 0,2.95=19(g)

mHCl=0,4.36,5=14,6(g) -> mddHCl= ?? Không cho C% sao tính ta