\(a/ CuO+2HCl \to CuCl_2+H_2O\\ b/\\ n_{CuO}=0,125(mol)\\ \to n_{HCl}=0,125.2=0,25(mol)\\ m_{HCl}=0,25.36,5=9,125(g)\\ c/\\ n_{CuO}=n_{CuCl_2}=0,125(mol)\\ CM_{CuCl_2}=\frac{0,125}{0,5}=0,25M\)

a) \(CuO+2HCl\rightarrow CuCl2+H2O\)

b) Ta có: \(n_{CuO}=\dfrac{10}{80}=0,8\left(mol\right)\)

Theo PT: \(n_{HCl}=2nCuO=1,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=1,6.36,5=58,4\left(g\right)\)

c) \(n_{CuCl2}=n_{CuO}=0,8\left(mol\right)\)

\(V_{dd}=\)không đổi \(=500ml=0,5l\)

\(\Rightarrow C_{M\left(CuCl2\right)}=\dfrac{0,8}{0,5}=1,6\left(M\right)\)

Bài 9 : 

\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

0,05--->0,1-------->0,05 

a) \(C_{MddHCl}=\dfrac{0,1}{0,1}=1\left(M\right)\)

b) \(m_{CuCl2}=0,05.135=6,75\left(g\right)\)

c) \(C_{MCuCl2}=\dfrac{0,05}{0,1}0,5\left(M\right)\)

Câu 10 : 

\(n_{FeO}=\dfrac{3,6}{72}=0,05\left(mol\right)\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

0,05-->0,1------->0,05

\(m_{ddHCl}=\dfrac{0,1.36,5}{10\%}100\%=36,5\left(g\right)\)

\(m_{ddspu}=3,6+36,5=40,1\left(g\right)\)

\(C\%_{FeCl2}=\dfrac{0,05.127}{40,1}.100\%=15,84\%\)

Câu 10 :

\(n_{FeO}=\dfrac{3,6}{72}=0,05\left(mol\right)\)

PTHH :

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

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

\(m_{ddHCl}=\dfrac{0,1.36,5.100}{10}=36,5\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{0,05.127}{\left(3,6+36,5\right)}.100\%=15,84\%\)

a, \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)

b, \(n_{CO_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{CO_2}=0,08\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,08}{0,2}=0,4\left(M\right)\)

c, \(n_{Na_2CO_3}=0,04\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{0,04.106}{10}.100\%=42,4\%\\\%m_{NaCl}=57,6\%\end{matrix}\right.\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)  
          0,2       0,4           0,2 
\(m_{MgCl_2}=0,2.95=19\left(g\right)\\ C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

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

a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

           1          2               1        1

         0,2       0,4            0,2       0,2

b) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)⁰/₀

c) \(n_{ZnCl2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)

⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)

\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)

\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)⁰/₀

 Chúc bạn học tốt

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

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

\(0.2........0.4..........0.2.......0.2\)

\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)

\(C\%_{HCl}=\dfrac{14.6}{100}\cdot100\%=14.6\%\)

\(m_{ZnCl_2}=0.2\cdot136=27.2\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=13+100-0.2\cdot2=112.6\left(g\right)\)

\(C\%_{ZnCl_2}=\dfrac{27.2}{112.6}\cdot100\%=24.1\%\)

nCuO=0,125mol

nH2SO4=0,5 mol

CuO +        2H2SO4-->     CuCl2 +   H2O

0,125          0,5

0,125          0,25                  0,125

C_CuCl2=0,125/0,5=0,25 M

C_{HCl du}=0,25/0,5=0,5 M

0                  0,25                  0,125

C_CuCl2= 0,125/

a) PTHH: \(KOH+HCl\rightarrow KCl+H_2O\)

b) Ta có: \(n_{KCl}=0,15\cdot0,5=0,075\left(mol\right)=n_{KOH}\) \(\Rightarrow m_{KOH}=0,075\cdot56=4,2\left(g\right)\)

c) PTHH: \(KCl+AgNO_3\rightarrow KNO_3+AgCl\downarrow\)

Theo PTHH: \(n_{KCl}=0,075\left(mol\right)=n_{AgNO_3\left(p.ứ\right)}=n_{KNO_3}=n_{AgCl}\)

\(\Rightarrow n_{AgNO_3\left(dư\right)}=0,075\cdot120\%-0,075=0,015\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{AgCl}=0,075\cdot143,5=10,7625\left(g\right)\\C_{M_{KNO_3}}=\dfrac{0,075}{0,5+2}=0,03\left(M\right)\\C_{M_{AgNO_3\left(dư\right)}}=\dfrac{0,015}{2,5}=0,006\left(M\right)\end{matrix}\right.\)

d) Coi như khi cô cạn không bị hao hụt muối

Ta có: \(m_{muối.khan}=m_{KNO_3}+m_{AgNO_3\left(dư\right)}=0,075\cdot101+0,015\cdot170=10,125\left(g\right)\)

\(n_{SO_3}=\dfrac{8}{80}=0.1\left(mol\right)\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(0.1......................0.1\)

\(C_{M_{H_2SO_4}}=\dfrac{0.1}{0.5}=0.2\left(M\right)\)

\(n_{CuO}=\dfrac{10}{80}=0.125\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(0.1...........0.1\)

\(m_{CuO\left(dư\right)}=\left(0.125-0.1\right)\cdot80=2\left(g\right)\)

\(a) 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ b)n_{AlCl_3} = n_{Al} = \dfrac{8,1}{27} = 0,3(mol)\\ \Rightarrow m_{AlCl_3} = 0,3.133,5 = 40,05(gam)\\ c) n_{HCl} = 3n_{Al} = 0,9(mol)\\ \Rightarrow C_{M_{HCl}} = \dfrac{0,9}{0,5} = 1,8M\)