nHCl=0,15 . 2=0,3(mol)

PTHH: CuO + 2 HCl -> CuCl2 + H2O

x___________2x____x(mol)

ZnO +2 HCl -> ZnCl2 + H2O

y______2y_____y(mol)

Ta có hpt: \(\left\{{}\begin{matrix}80x+81y=12,1\\2x+2y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)

Vddsau=VddHCl=0,15(l)

=> CMddCuO= 0,05/0,15=1/3(M)

CMddZnO=0,1/0,15=2/3(M)

a) Đặt nCuO=x(mol); nZnO=y(mol) (x,y>0)

nHCl=0,3(mol)

PTHH: CuO +2  HCl -> CuCl2 + H2O

x_________2x______x(mol)

ZnO +2 HCl -> ZnCl2 + H2O

y_____2y_____y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}80x+81y=12,1\\2x+2y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)

b) mCuO= 80x=4(g)

=>%mCuO=(4/12,1).100=33,058%

=>%mZnO=66,942%

C) mH2SO4=98.0,3=29,4(g)

=>mddH2SO4=(29,4.100)/20=147(g)

a/ Gọi x,y lần lượt là số mol CuO và ZnO tham gia phản ứng

nHCl = 14,6/36,5 = 0,4 (mol)

PTHH : CuO + 2HCl -----> CuCl₂ + H₂O

 (mol)      x         2x               x

             ZnO + 2HCl -----> ZnCl₂ + H₂O

(mol)        y        2y               y

Ta có hệ pt : \(\begin{cases}80x+81y=16,08\\2x+2y=0,4\end{cases}\) \(\Leftrightarrow\begin{cases}x=0,12\\y=0,08\end{cases}\)

=> mCuO = 0,12.80 = 9,6 (g)

\(\Rightarrow\%CuO=\frac{9,6}{16,08}.100\approx59,7\%\)

=> %ZnO = 100% - 59,7% = 40,3%

b/ mCuCl₂ = 0,12.135 = 16,2(g)

mZnCl₂ = 0,08.136 = 10,88 (g)

B1:

2NaOH+H₂SO₄\(\rightarrow\)Na₂SO₄+2H₂O

n_NaOH=\(\frac{4}{40}=0.1\)mol

=>n_H2SO4=\(\frac{1}{2}\)n_NaOH=0.05 mol

=>C_M=\(\frac{n_{H2SO42}}{V}\)=\(\frac{0.05}{200}\)=2,5.10^-4 (M)

B2:

Mg+\(\frac{1}{2}\)O₂\(\underrightarrow{t^0}\)MgO                (1)

MgO+2HCl\(\rightarrow\)MgCl₂+H₂O (2)

n_Mg(1)=\(\frac{0,36}{24}=0,015mol\)

=>n_MgO(1)=0,015=n_MgO(2)

n_HCl(2)=2n_MgO(2)=0,03mol

=>C_M(HCl)=\(\frac{n_{HCl}}{V}=\frac{0,03}{100}=3.10^{-4}M\)

\(a)n_{HCl}=0,1.3=0,3mol\\ n_{CuO}=a;n_{ZnO}=b\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ ZnO+2HCl\rightarrow ZnCl_2+H_2O\\ \Rightarrow\left[{}\begin{matrix}80a+81b=12,1\\2a+2b=0,3\end{matrix}\right.\\ \Rightarrow a=0,05;b=0,1\\ \%m_{CuO}=\dfrac{80.0,05}{12,1}\cdot100=33,06\%\\ \%m_{ZnO}=100-33,06=66,94\%\\ b)ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O \\ n_{H_2SO_4}=0,05+0,1=0,15mol\\ m_{ddH_2SO_4}=\dfrac{0,15.98}{20}\cdot100=73,5g\)

Ta có: \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(n_{HCl}=0,4.1,5=0,6\left(mol\right)\)

Giả sử: \(\left\{{}\begin{matrix}n_{Na_2CO_3}=x\left(mol\right)\\n_{K_2CO_3}=y\left(mol\right)\end{matrix}\right.\)

PT: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)

\(K_2CO_3+2HCl\rightarrow2KCl+H_2O+CO_2\)

\(HCl_{dư}+NaOH\rightarrow NaCl+H_2O\)

Theo PT: \(n_{CO_2}=n_{Na_2CO_3}+n_{K_2CO_3}=x+y\left(mol\right)\) ⇒ x + y = 0,25 (1)

\(n_{HCl\left(pư\right)}=2x+2y\left(mol\right)\) \(\Rightarrow n_{HCl\left(dư\right)}=0,6-2x-2y\left(mol\right)\)

Có: \(\left\{{}\begin{matrix}n_{NaCl}=2n_{Na_2CO_3}+n_{HCl\left(dư\right)}=0,6-2y\left(mol\right)\\n_{KCl}=2n_{K_2CO_3}=2y\left(mol\right)\end{matrix}\right.\)

⇒ 58,5(0,6 - 2y) + 74,5.2y = 39,9 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{0,1.106}{0,1.106+0,15.138}.100\%\approx33,9\%\\\%m_{K_2CO_3}\approx66,1\%\end{matrix}\right.\)

Bạn tham khảo nhé!