\(n_O=\dfrac{21,2-14}{16}=0,45\left(mol\right)\)

=> n_H2O = 0,45 (mol)

Giả sử có V (lít) dd Z

=> \(\left\{{}\begin{matrix}n_{HCl}=0,1V\left(mol\right)\\n_{H_2SO_4}=0,1V\left(mol\right)\end{matrix}\right.\)

Bảo toàn H: 0,1V + 2.0,1V = 0,45.2

=> V = 3 (lít)

a,\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,2      0,4                       0,2

\(\Rightarrow\%m_{Zn}=\dfrac{0,2.65.100\%}{21,1}=61,61\%;\%m_{ZnO}=100-61,61=38,39\%\)

b,\(n_{ZnO}=\dfrac{21,1-13}{81}=0,1\left(mol\right)\)

PTHH: ZnO + 2HCl → ZnCl₂ + H₂O

Mol:     0,1       0,2         

\(m_{ddHCl}=\dfrac{\left(0,2+0,4\right).36,5.100\%}{7,3\%}=300\left(g\right)\)

c,

PTHH: Zn + H₂SO₄ → ZnSO₄ + H₂

Mol:     0,2       0,2

PTHH: ZnO + H₂SO₄ → ZnSO₄ + H₂O

Mol:      0,1        0,1

\(n_{H_2SO_4}=0,2+0,1=0,3\left(mol\right)\Rightarrow V_{ddH_2SO_4}=\dfrac{0,3}{0,5}=0,6\left(l\right)=600\left(ml\right)\)

\(m_{ddH_2SO_4}=600.1,12=672\left(g\right)\)

a) Gọi số mol Mg, CuO là a, b (mol)

=> 24a + 80b = 14 (1)

\(n_{HCl}=\dfrac{255,5.10\%}{36,5}=0,7\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

             a--->2a--------->a------>a

           CuO + 2HCl --> CuCl₂ + H₂O

               b------>2b----->b

=> 2a + 2b = 0,7 (2)

(1)(2) => a = 0,25 (mol); b = 0,1 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,25.24}{14}.100\%=42,857\%\\\%m_{CuO}=\dfrac{0,1.80}{14}.100\%=57,143\%\end{matrix}\right.\)

b)

m_{dd sau pư} = 14 + 255,5 - 0,25.2 = 269 (g)

\(C\%_{MgCl_2}=\dfrac{0,25.95}{269}.100\%=8,829\%\)

\(C\%_{CuCl_2}=\dfrac{0,1.135}{269}.100\%=5,019\%\)

\(Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O\\ ZnO + 2HCl \to ZnCl_2 + H_2O\\ MgO + 2HCl \to MgCl_2 + H_2O\\ n_{HCl} = 0,5.0,1 = 0,05(mol)\\ n_{H_2O} = \dfrac{1}{2}n_{HCl} = 0,025(mol)\\ \text{Bảo toàn khối lượng : }\\ m_{muối} = m_{hỗn\ hợp} + m_{HCl} - m_{H_2O} = 1,405 + 0,05.36,5 - 0,025.18 = 2,78(gam)\)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ Mg+H_2SO_4\to MgSO_4+H_2\\ MgO+H_2SO_4\to MgSO_4+H_2O\\ \Rightarrow n_{Mg}=0,25(mol)\\ a,\begin{cases} \%_{Mg}=\dfrac{0,25.24}{14}.100\%=42,86\%\\ \%_{MgO}=100\%-42,86\%=57,14\% \end{cases}\\ b,n_{MgO}=\dfrac{14-0,25.24}{40}=0,2(mol)\\ \Rightarrow \Sigma n_{H_2SO_4}=0,2+0,25=0,45(mol)\\ \Rightarrow C\%_{H_2SO_4}=\dfrac{0,45.98}{200}.100\%=22,05\%\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)  (1)

             \(MgO+2HCl\rightarrow MgCl_2+H_2O\)  (2)

a) Ta có: \(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)=n_{Mg}\) \(\Rightarrow m_{Mg}=1\cdot24=24\left(g\right)\)

\(\Rightarrow\%m_{Mg}=\dfrac{24}{32}\cdot100\%=75\%\) \(\Rightarrow\%m_{MgO}=25\%\)

b) Theo 2 PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2n_{Mg}=2mol\\n_{HCl\left(2\right)}=2n_{MgO}=2\cdot\dfrac{32-24}{40}=0,4mol\end{matrix}\right.\)

\(\Rightarrow\Sigma n_{HCl}=2,4mol\) \(\Rightarrow m_{ddHCl}=\dfrac{2,4\cdot36,5}{7,3\%}=1200\left(g\right)\)

c) Theo PTHH: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=1,2mol\)

\(\Rightarrow\Sigma m_{MgCl_2}=1,2\cdot95=114\left(g\right)\)

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

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=1230\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{114}{1230}\cdot100\%\approx9,27\%\)

_{Pư: Mg+2HCl -> MgCl2 +H2}

_{MgO+2HCl-> MgCl2 +H2O}

_{MgCl2+2NaOH-> Mg(OH)2}↓_+2NaCl

_{Mg(OH)2-> MgO+H2O}

nMgO=14/40=0.35 mol

-> nMg(OH)2=nMgCl2=nMgO=0.35 mol

-> nMgO=nMg=nMgCl2=0.35 mol

gọi x,y lần lượt là số mol của Mg,MgO

->mhh=24x+40y=2 (1)

nMgCl2=x+y=0.35 (2)

giải hệ (1)(2) có x=0.103125

y=0.246875

- khối lượng của bạn nhập sai vì klg hỗn hợp quá nhỏ

ông tusho sjdo judo viết Hệ PT sai . tôi bấm  mấy PT ông này y ra âm

Đặt: nMgO=x(mol) ; nCuO=y(mol) (x,y>0)

nHCl=0,2.2=0,4(mol)

PTHH: MgO +2 HCl -> MgCl2 + H2

x________2x__________x___x(mol)

CuO +2 HCl -> CuCl2 + H2

y___2y______y_____y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}40x+80y=10\\2x+2y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,05\end{matrix}\right.\)

=> mMgO= 0,15.40=6(g); mCuO= 0,05.80=4(g)