\(n_{CuO}=\dfrac{10}{80}=0,125\left(mol\right)\)

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

\(\Rightarrow n_{HCl}=2n_{CuO}=0,25\left(mol\right)\)

\(\Rightarrow V_{dd}=\dfrac{0,25}{3}=\dfrac{1}{12}\left(l\right)\)

\(n_{CuCl_2}=n_{CuO}=0,125\left(mol\right)\)

\(\Rightarrow C_{M\left(CuCl_2\right)}=\dfrac{0,125}{\dfrac{1}{12}}=1,5M\)

a)

nFe2O3=16/160=0,1(mol)

nHCl=0,5.1=0,5(mol)

PTHH: Fe2O3 + 6 HCl -> 2 FeCl3 + 3 H2O

Ta có: 0,1/1 > 0,5/6

=> HCl hết, Fe2O3 dư, tính theo nHCl.

nFeCl3= 2/6. nHCl= 2/6 . 0,5= 1/6(mol)

=>mFeCl3= 162,5. 1/6= 27,083(g)

b) Vddsau=VddHCl=0,5(l)

- dd sau p.ứ chỉ có FeCl3.

=> CMddFeCl3= 1/6: 0,5= 1/3(M)

PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\\n_{HCl}=0,5\cdot0,5=0,25\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,25}{2}\) \(\Rightarrow\) HCl còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{CaCl_2}=0,1\left(mol\right)\\n_{HCl\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{CaCl_2}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,05}{0,5}=0,1\left(M\right)\end{matrix}\right.\)

a) 2Al + 6HCl --> 2AlCl₃ + 3H₂

b)

\(n_{Al}=\dfrac{6,75}{27}=0,25\left(mol\right)\); \(n_{HCl}=0,6.1=0,6\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,25}{2}>\dfrac{0,6}{6}\) => Al dư, HCl hết

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             0,2<--0,6------------->0,3

=> V = 0,3.22,4 = 6,72 (l)

c) \(m_{Al\left(dư\right)}=\left(0,25-0,2\right).27=1,35\left(g\right)\)

d) \(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

Xét tỉ lệ: \(\dfrac{0,25}{1}< \dfrac{0,3}{1}\) => CuO hết

PTHH: CuO + H₂ --t^o--> Cu + H₂O

              0,25---------->0,25

=> m_Cu = 0,25.64 = 16 (g)

a) 2Al + 6HCl --> 2AlCl₃ + 3H₂

b)

\(n_{Al}=\dfrac{6,75}{27}=0,25\left(mol\right)\); \(n_{HCl}=0,6.1=0,6\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,25}{2}>\dfrac{0,6}{6}\) => Al dư, HCl hết

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

                      0,6-------------->0,3

=> V = 0,3.22,4 = 6,72 (l)

d) \(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

Xét tỉ lệ: \(\dfrac{0,25}{1}< \dfrac{0,3}{1}\) => CuO hết

PTHH: CuO + H₂ --t^o--> Cu + H₂O

              0,25---------->0,25

=> m_Cu = 0,25.64 = 16 (g)

 Xem hình!

a) $CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$

n CaCO3 = n CO2 = x(mol)

m giảm = m CaCO3 - m CO2 

=> 3,36 = 100x -44x

=> x = 0,06(mol)

Gọi n MgCO3 = a(mol) ; n CaCO3 = b(mol)

=> 84a + 100b = 5,68(1)

$MgCO_3 + 2HCl \to MgCl_2 + CO_2 + H_2O$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
n CO2 = a + b = x = 0,06(2) 

Từ (1)(2) suy ra a = 0,02 ; b = 0,04

%m MgCO3 = 0,02.84/5,68   .100% = 29,58%
%m CaCO3 = 100% -29,58% = 70,42%

b)

n MgCl2 = a = 0,02(mol)

n CaCl2 = b = 0,04(mol)

n HCl pư = 2a + 2b = 0,12(mol)

=> n HCl dư = 0,5.1 - 0,12 = 0,38(mol)

Vậy : 

CM MgCl2 = 0,02/1 = 0,02M

CM CaCl2 = 0,04/1 = 0,04M

CM HCl = 0,38/1 = 0,38M

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

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

b, Gọi: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\) ⇒ 80x + 160y = 32 (1)

Theo PT: \(\left\{{}\begin{matrix}n_{CuCl_2}=n_{Cu}=x\left(mol\right)\\n_{FeCl_3}=2n_{Fe_2O_3}=2y\left(mol\right)\end{matrix}\right.\) ⇒ 135x + 325y = 59,5 (2)

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

\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2.80=16\left(g\right)\\m_{Fe_2O_3}=0,1.160=16\left(g\right)\end{matrix}\right.\)

c, Theo PT: \(n_{HCl}=2n_{CuO}+6n_{Fe_2O_3}=1\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{1}{0,5}=2\left(l\right)\)

Câu 1 : 

\(n_{Mg}=\dfrac{8.4}{24}=0.35\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0.35.......0.7.........0.35..........0.35\)

\(C\%_{HCl}=\dfrac{0.7\cdot36.5}{146}\cdot100\%=17.5\%\)

\(m_{\text{dung dịch sau phản ứng}}=8.4+146-0.35\cdot2=153.7\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{0.35\cdot95}{153.7}\cdot100\%=21.6\%\)

Câu 2 :

\(n_{CaCO_3}=\dfrac{10}{100}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{114.1\cdot8\%}{36.5}=0.25\left(mol\right)\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

\(1................2\)

\(0.1.............0.25\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.25}{2}\Rightarrow HCldư\)

\(m_{\text{dung dịch sau phản ứng}}=10+114.1-0.1\cdot44=119.7\left(g\right)\)

\(C\%_{HCl\left(dư\right)}=\dfrac{\left(0.25-0.2\right)\cdot36.5}{119.7}\cdot100\%=1.52\%\)

\(C\%_{CaCl_2}=\dfrac{0.2\cdot111}{119.7}\cdot100\%=18.54\%\)

Câu 3 : 

\(n_{H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.3.......0.6......................0.3\)

\(m_{Fe}=0.3\cdot56=16.8\left(g\right)\)

\(m_{dd_{HCl}}=150\cdot1.2=180\left(g\right)\)

\(C\%_{HCl}=\dfrac{0.6\cdot36.5}{180}\cdot100\%=12.16\%\)

\(C_{M_{HCl}}=\dfrac{0.6}{0.15}=4\left(M\right)\)