a,Fe     +        2HCl            →            FeCl               +              H_{2           (1)}

   FeO   +        2HCl            →            FeCl               +              H₂O       (2)

n_H2 =  3,36/ 22,4 = 0,15 ( mol)

Theo (1)  n_H2 = nFe =  0,15 ( mol)

m_Fe = 0,15 x 56  =  8.4 (g)

m _FeO = 12 - 8,4  =  3,6 (g)

a, \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)  

\(Fe+2HCl->FeCl_2+H_2\left(1\right)\) 

\(FeO+2HCl->FeCl_2+H_2O\left(2\right)\) 

theo (1) \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\) 

=> \(m_{Fe}=0,15.56=8,4\left(g\right)\) 

=> \(m_{FeO}=12-8,4=3,6\left(g\right)\)

ta thấy : nFe =nH2 = 0,15

=> mFe =0,15 x 56 = 8,4g

%Fe=8,4/12 x 100 = 70%

=>%FeO = 100 - 70 = 30%

b) BTKLra mdd tìm mct of HCl

c) tìm mdd sau pứ -mH2 nha bạn

a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

                   a_____2a______a_____a      (mol)

                \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

                    b_____3b_______b_____\(\dfrac{3}{2}\)b         (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}56a+27b=36,1\\a+\dfrac{3}{2}b=\dfrac{21,28}{22,4}=0,95\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,5\\b=0,3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,5\cdot56=28\left(g\right)\\m_{Al}=8,1\left(g\right)\end{matrix}\right.\)

b+c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{HCl}=2a+3b=1,9\left(mol\right)\\n_{FeCl_2}=0,5\left(mol\right)\\n_{AlCl_3}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{1,9}{0,2}=9,5\left(M\right)\\C_{M_{FeCl_2}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\)

\(a)2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b)n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\\ n_{Al}=a;n_{Fe}=b\\ \left\{{}\begin{matrix}3a+b=0,25\\27a+56b=8,3\end{matrix}\right.\\ a=\dfrac{19}{470};b=\dfrac{121}{940}\\ \%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{8,3}\cdot100=13,15\%\\ \%m_{Fe}=100-13,15=86,85\%\\ c)n_{HCl}=3\cdot\dfrac{19}{470}+2\cdot\dfrac{121}{940}=\dfrac{89}{235}mol\\ m_{ddHCl=}=\dfrac{\dfrac{89}{235}\cdot36,5}{7,3}\cdot100=189g\\ d)n_{AlCl_3}=n_{Al}=\dfrac{19}{470}mol\\ n_{Fe}=n_{FeCl_2}=\dfrac{121}{940}mol\)

\(m_{dd}=8,3+189-0,25.2=196,8g\\ C_{\%AlCl_3}=\dfrac{\dfrac{19}{470}\cdot133,8}{196,8}\cdot100=2,8\%\\ C_{\%FeCl_2}=\dfrac{\dfrac{121}{940}127}{196,8}\cdot100=8,3\%\)

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

Theo PT: \(n_{Mg}=n_{H_2}=0,03\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,03.24=0,72\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,72}{1,74}.100\%\approx41,38\%\\\%m_{AlCl_3}\approx58,62\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=0,06\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,06.36,5=2,19\left(g\right)\)

\(\Rightarrow C\%_{ddHCl}=\dfrac{2,19}{500}.100\%=0,438\%\)

Bạn tham khảo nhé!

a, nH2 = 0,03 ( mol )

=> nMg = nH2 = 0,03 ( mol )

=> mMg = 0,72 g

=> %Mg \(\approx\) 41,38 % .

=> % Al \(\approx\) 58,62 % .

b, Có : nH2 = 0,03 mol

=> nHCl = nHCl_{từ Al2O3} + nHCl_{từ Mg} = 0,06 + 0,06 = 0,12 ( mol )

=> mHCl = 4,38 ( g )

Lại có : mdd = mhh + mddHCl = 501,74 ( g )

=> \(C\%=\dfrac{m_{HCl}}{m_{dd}}.100\%\approx0,87\%\)

( chắc đoạn trên là Al2O3 :vvvv )

a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,5.65=32,5\left(g\right)\)

\(\Rightarrow m_{CuO}=72,5-32,5=40\left(g\right)\)

c, Ta có: \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Zn}+n_{CuO}=1\left(mol\right)\)

\(\Rightarrow b=C_{M_{H_2SO_4}}=\dfrac{1}{2,5}=0,4M\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnSO_4}=n_{Zn}=0,5\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnSO_4}}=\dfrac{0,5}{2,5}=0,2M\\C_{M_{CuSO_4}}=\dfrac{0,5}{2,5}=0,2M\end{matrix}\right.\)

Bạn tham khảo nhé!

a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\left(I\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\left(II\right)\)

b, Theo PTHH(1) : \(n_{Zn}=n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{Zn}=32,5\left(g\right)\)

\(\Rightarrow m_{CuO}=m_{hh}-m_{Zn}=40\left(g\right)\)

\(\Rightarrow n_{CuO}=\dfrac{m}{M}=0,5\left(mol\right)\)

c, Theo PTHH (1) và (2) : \(n_{H2SO4}=n_{CuO}+n_{Zn}=1\left(mol\right)\)

\(\Rightarrow C_{MH2SO4}=b=\dfrac{n}{V}=\dfrac{1}{2,5}=0,4M\)

d, ( Chắc là thể tích coi như không đổi )

Thấy sau phản ứng thu được A gồm \(0,5molZnSO_4,0,5molCuSO_4\)

\(\Rightarrow C_{MCuSO4}=C_{MZnSO4}=\dfrac{n}{V}=\dfrac{0,5}{2,5}=0,2M\)

Vậy ...

CuO + 2HCl -> CuCl₂ + H₂O (1)

Fe₂O₃ + 6HCl -> 2FeCl₃ + 3H₂O (2)

n_HCl=0,2.3,5=0,7(mol)

Đặt n_CuO=a

n_Fe2O3=b

Ta có hệ:

80a+160b=20

2a+6b=0,7

=>a=0,05;b=0,1

m_CuO=80.0,05=4(g)

m_Fe2O3=20-4=16(g)

Theo PTHH 1 và 2 ta có:

n_CuCl2=n_CuO=0,05(mol)

n_FeCl3=2n_Fe2O3=0,2(mol)

m_CuCl2=135.0,05=6,75(g)

m_FeCl3=162,5.0,2=32,5(g)

m_dd =20+200.1,1=240(g)

C% dd CuCl₂=6,72\240 .100%=2,8125%

C% dd FeCl₃= 32,5\240 .100%=13,54%