\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,1           0,2            0,1      0,1 
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\ V_{H_2}=0,1.22,4=2,24l\\ m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)  
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,2             0,4       0,2              0,2 
\(m_{HCl}=0,4.36,5=14,6g\\ V_{H_2}=0,2.22,4=4,48l\\ m\text{dd}=4,8+200-0,4=204,4g\\ C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)

\(n_{Mg}=2x\left(mol\right),n_{Fe}=x\left(mol\right)\)

\(n_{HCl}=0.2\cdot0.45=0.9\left(mol\right)\)

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

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

\(n_{HCl}=2\cdot2x+2\cdot x=0.9\left(mol\right)\)

\(\Rightarrow x=0.15\)

\(m_{hh}=0.3\cdot24+0.15\cdot56=15.6\left(g\right)\)

\(V_{H_2}=0.45\cdot22.4=10.08\left(l\right)\)

Bài 1 : 

\(a) Fe_2O_3 + 3H_2 \xrightarrow{t^o}2Fe + 3H_2O\\ b) n_{Fe_2O_3} = \dfrac{80}{160}= 0,5(mol)\\ n_{H_2} = 3n_{Fe_2O_3} = 1,5(mol)\\ \Rightarrow V_{H_2} = 1,5.22,4 = 33,6(lít)\\ n_{Fe} = 2n_{Fe_2O_3} = 1(mol)\\ m_{Fe} = 1.56 = 56(gam)\)

Bài 2 :

\(a) Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = n_{Fe} =\dfrac{5,6}{56} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ n_{HCl} =2 n_{Fe} = 0,2(mol)\\ m_{HCl} = 0,2.36,5 = 7,3(gam)\)

Zn+2HCl→ZnCl2+H2

 2Al+6HCl→2AlCl3+3H2

Gọi số mol HCllà x

n H2=\(\dfrac{1}{2}\)n HCl=0,5x

Bảo toàn khối lượng: mkl+mHCl=mmuối+mH2

6,05+36,5x=13,15+0,5x.2

→x=0,2 mol

mHCl=0,2.36,5=7,3 gam

=>m =mddHCl=\(\dfrac{7,3}{10\%}\)=73gam

Câu 1:

Gọi : nMg=a(mol); nMgO=b(mol) (a,b>0)

a) PTHH: Mg + 2 HCl -> MgCl2 + H2

a________2a_______a______a(mol)

MgO +2 HCl -> MgCl2 + H2O

b_____2b_______b___b(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}24a+40b=8,8\\22,4a=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

=> mMg=0,2.24=4,8(g)

=>%mMg= (4,8/8,8).100=54,545%

=> %mMgO= 45,455%

b) m(muối)=mMg2+ + mCl- = 0,3. 24 + 0,6.35,5=28,5(g)

c) V=VddHCl=(2a+2b)/2=0,3(l)=300(ml)

Câu 2:

Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{HCl}=0,4\cdot2=0,8\left(mol\right)\end{matrix}\right.\)

PTHH: \(Ca+2HCl\rightarrow CaCl_2+H_2\uparrow\)

               0,2____0,4_____0,2____0,2   (mol)

           \(CaO+2HCl\rightarrow CaCl_2+H_2O\)

                0,2____0,4______0,2____0,2  (mol)

Ta có: \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,2\cdot40}{0,2\cdot40+0,2\cdot56}\cdot100\%\approx41,67\%\\\%m_{CaO}=58,33\%\\m_{CaCl_2}=\left(0,2+0,2\right)\cdot111=44,4\left(g\right)\end{matrix}\right.\)

\(Zn+2HCl->ZnCl_2+H_2\\ Mg+2HCl->MgCl_2+H_2\\ n_{Zn}=a\\ n_{Mg}=b\\ 65a+24b=11,3g\\ n_{H_2}=a+b=\dfrac{6,72}{22,4}=0,3\\ a=0,1\\ m_{Zn}=65.0,1=6,5g\)

\(n_{MnO_2}=\dfrac{3,48}{87}=0,04(mol)\\ n_{Cl_2}=\dfrac{0,672}{22,4}=0,03(mol)\\ a,MnO_2+4HCl\xrightarrow{t^o}MnCl_2+Cl_2+2H_2O\\ \Rightarrow n_{Cl_2(p/ứ)}=0,04(mol)\\ \Rightarrow V_{Cl_2(p/ứ)}=0,04.22,4=0,896(l)\\ \Rightarrow H\%=\dfrac{0,672}{0,896}.100\%=75\%\)

\(b,n_{Cu}=\dfrac{0,64}{64}=0,01(mol)\\ 2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3(1)\\ Cu+Cl_2\xrightarrow{t^o}CuCl_2(2)\\ \Rightarrow n_{Cl_2(2)}=n_{Cu}=0,01(mol)\\ \Rightarrow n_{Cl_2(1)}=0,03-0,01=0,02(mol)\\ \Rightarrow n_{Fe}=\dfrac{1}{75}(mol) \Rightarrow m_{Fe}=\dfrac{1}{75}.56=0,75(g)\\ \Rightarrow m_{hh}=0,75+0,64=1,39(g)\)

\(c,FeCl_3+3AgNO_3\to Fe(NO_3)_3+3AgCl\downarrow\\ CuCl_2+2AgNO_3\to Cu(NO_3)_2+2AgCl\downarrow\\ \Rightarrow \Sigma n_{AgCl}=3n_{FeCl_3}+2n_{CuCl_2}=0,04+0,01=0,05(mol)\\ \Rightarrow m_{\downarrow}=\Sigma m_{AgCl}=0,05.143,5=7,175(g)\\ d,FeCl_3+3NaOH\to Fe(OH)_3\downarrow+3NaCl\\ CuCl_2+2NaOH\to Cu(OH)_2\downarrow+2NaCl\\ \Rightarrow \Sigma n_{naOH}=3n_{FeCl_3}+2n_{CuCl_2}=0,05(mol)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,05}{2}=0,025(mol)\\ \Rightarrow m_{dd_{NaOH}}=0,025.1,12=0,028(g)\)

a)MnO₂+4HCl→MnCl₂+Cl₂+2H₂O

nMnO₂=3,48/87=0,04mol

n_Cl2=n_MnO2=0,04mol

V_Cl2=0,04×22,4=0,896l

H=0,672/0,896×100%=75%

b)2Fe+3Cl₂→2FeCl₃(1)

Cu+Cl₂→CuCl₂

n_Cu=0,64/64=0,01mol

n_CuCl2=n_Cu=0,01mol

m_CuCl2=0,01×135=1,35g

n_Cl2(1)=0,03−0,01=0,02mol

n_FeCl3=2/3.n_Cl2(1)=1/75mol

m_FeCl3=1/75×162,5=2,17g

m=1,35+2,17=3,52g