Ta có n_HCl = \(\dfrac{10,95}{36,5}\) = 0,3 ( mol )

CuO + 2HCl \(\rightarrow\) CuCl₂ + H₂O

x → 2x → x → x

Zno + 2HCl \(\rightarrow\) ZnCl₂ + H₂O

y → 2y → y → y

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

=> % CuO = \(\dfrac{64\times0.05}{12,1}\) . 100 \(\approx\) 26,5%

=> % ZnO = 100 - 26,5 = 73,5 %

=> M_H2O = ( x + y ).18 = ( 0,05 + 0,1 ).18 = 2,7 (gam)

a) Gọi `n_{CuO} = a (mol); n_{ZnO} = b (mol)`

`=> 80a + 81b = 12,1(1)`

`n_{HCl} = 3.0,1 = 0,3 (mol)`

PTHH:
`CuO + 2HCl -> CuCl_2 + H_2O`

`ZnO + 2HCl -> ZnCl_2 + H_2O`

Theo PT: `n_{HCl} = 2n_{ZnO} + 2n_{CuO}`

`=> 2a + 2b = 0,3 (2)`

Từ `(1), (2) => a = 0,05; b = 0,1`

`=> \%m_{CuO} = (0,05.80)/(12,1) .100\% = 33,06\%`

`=> \%m_{ZnO} = 100\% - 33,06\% = 66,94\%`

b) PTHH:
`CuO + H_2SO_4 -> CuSO_4 + H_2O`

`ZnO + H_2SO_4 -> ZnSO_4 + H_2O`

Theo PT: `n_{H_2SO_4} = n_{CuO} + n_{ZnO} = 0,15 (mol)`

`=> m_{ddH_2SO_4} = (0,15.98)/(20\%) = 73,5 (g)`

\(a.CuO+2HCl\rightarrow CuCl_2+H_2O\\ ZnO+2HCl\rightarrow ZnCl_2+H_2\\ n_{HCl}=0,4.3=1,2mol\\ n_{CuCl_2}=a;n_{ZnCl_2}=b\\ \Rightarrow\left\{{}\begin{matrix}80a+81b=48,5\\2a+2b=1,2\end{matrix}\right.\\ \Rightarrow a=0,1,b=0,5\\ m_O=16\left(0,1+0,5\right)=9,6g\\ c.m_{CuCl_2}=0,1.135=13,5g\\ m_{ZnCl_2}=0,5.136=68g\\ d.C_{M_{CuCl_2}}=\dfrac{0,1}{0,4}=0,25M\\ C_{M_{ZnCl_2}}=\dfrac{0,5}{0,4}=1,25M\)

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

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

b, \(n_{HCl}=0,4.3=1,2\left(mol\right)\)

→ \(O+2H\rightarrow H_2O\)

⇒ n_O = 0,6 (mol)

⇒ m_O = 0,6.16 = 9,6 (g)

c, Theo PT: \(n_{H_2O}=\dfrac{1}{2}n_{HCl}=0,6\left(mol\right)\)

Theo ĐLBT KL, có: m _oxit + m_HCl = m _muối + m_H2O

⇒ 48,5 + 1,2.36,5 = m _muối + 0,6.18

⇒ m _muối = 81,5 (g)

d, Ta có: 135n_CuCl2 + 136n_ZnCl2 = 81,5 (1)

Theo PT: \(\left\{{}\begin{matrix}n_{CuO}=n_{CuCl_2}\\n_{ZnO}=n_{ZnCl_2}\end{matrix}\right.\) ⇒ 80n_CuCl2 + 81n_ZnCl2 = 48,5 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CuCl_2}=0,1\left(mol\right)\\n_{ZnCl_2}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{CuCl_2}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\C_{M_{ZnCl_2}}=\dfrac{0,5}{0,4}=1,25\left(M\right)\end{matrix}\right.\)

Đặt : \(n_{CuO}=a\left(mol\right),n_{ZnO}=b\left(mol\right)\)

\(\Rightarrow80a+81b=28,25g\left(1\right)\)

a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

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

b) Ta có : \(n_{HCl}=\dfrac{20\%.127,75}{100\%.36,5}=0,7\left(mol\right)\Rightarrow2a+2b=0.7\left(2\right)\)

Từ (1),(2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1=n_{CuCl2}\\b=0,25=n_{ZnCl2}\end{matrix}\right.\)

c) \(m_{muối}=m_{CuCl2}+m_{ZnCl2}=0,1.135+0,25.136=47,5\left(g\right)\)

d) \(\left\{{}\begin{matrix}C\%_{CuCl2}=\dfrac{0,1.135}{28,25+127,75}.100\%=8,65\%\\C\%_{ZnCl2}=\dfrac{0,25.136}{28,25+127,75}.100\%=21,79\%\end{matrix}\right.\)

a) nHCl= 0,625.0,8= 0,5(mol)

Đặt: nMgO=a(mol); nAl2O3=b(mol) (a,b>0)

Vì tổng số mol MgO và Al2O3 bằng 0,15 nên ta có pt:

a+b=0,15 (1)

PTHH: ZnO +2 HCl -> ZnCl2 + H2O

a_________2a________a(mol)

Al2O3 + 6 HCl -> 2 AlCl3 + 3 H2O 

b______6b_____2b(mol)

nHCl(tổng)= 0,5

<=>2a+6b=0,5 (2)

Từ (1), (2) ta có hpt:

\(\left\{{}\begin{matrix}a+b=0,15\\2a+6b=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)

b) Khối lượng từng muối tạo thành:

mZnCl2= 136a=136.0,1=13,6(g)

mAlCl3=133,5.2b=267.0,05=13,35(g)

a)

\(Zn + 2HCl \to ZnCl_2 + H_2\\ Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2\)

Theo PTHH : \(n_{Zn} = n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\)

\(\Rightarrow n_{Fe_2O_3} = \dfrac{35,5-0,3.65}{160} = 0,1\\ \Rightarrow n_{HCl} = 2n_{Zn} + 6n_{Fe_2O_3} = 0,3.2 + 0,1.6 = 1,2(mol)\\ \Rightarrow m_{HCl} = 1,2.36,5 = 43,8(gam)\)

b)

\(CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\)

Gọi \(n_{CuO} = a;n_{Fe_2O_3} = b\)

\(\left\{{}\begin{matrix}80a+160b=19,6\\a+3b=0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,135\\b=0,055\end{matrix}\right.\)

Vậy : 

\(\left\{{}\begin{matrix}n_{Cu}=0,135\\n_{Fe}=0,055.2=0,11\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Cu}=0,135.64=8,64\left(gam\right)\\m_{Fe}=0,11.56=6,16\left(gam\right)\end{matrix}\right.\)

gọi số mol của CuO là x ZnO là y mol 

\(CuO+2HCl->CuCl_2+H_2O\)

x...........2x

\(ZnO+2HCl->ZnCl_2+H_2O\)

y............2y

n HCl= 3.0,1=0,3 mol 

m CuO+ m ZnO = 80x + 81y =12,1

2x+2y=0,3 

x=0,05 mol , y=0,1 mol 

m CuO = 0,05.(64+16)=4g

m ZnO = 0,1.(65+16) =8,1 g