\(n_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)

Gọi số mol Fe và Fe2O3 là a và b

Ta có \(56a+160b=21,6\)

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

\(Fe_2O_3+6HCl\rightarrow2FeCl_2+3H_2\)

a. \(n_{Fe}=n_{H2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\%m_{Fe}=\frac{0,1.56}{21,6}=25,92\%\)

\(\%m_{Fe2O3}=100\%-25,92\%=74,08\%\)

b. \(n_{HCl}=2n_{Fe}+6n_{Fe2O4}=2.0,1+6.0,1=0,8\)

\(\Rightarrow m_{HCl}=0,8.36,5=29,2\)

\(m_{dd_{HCl}}=\frac{29,2}{14,6\%}=200\left(g\right)\)

Khối lượng dd sau phản ứng:

m=m rắn+mddHCl-mH2

\(\Rightarrow m=21,6+200-0,1.2=221,4\left(g\right)\)

\(C\%_{FeCl2}=\frac{0,1.127}{221,4}=5,74\%\)

\(C\%_{FeCl3}=\frac{0,2.162,5}{221,4}=14,68\%\)

\(Đặt:\left\{{}\begin{matrix}Fe:x\left(mol\right)\\Zn:y\left(mol\right)\end{matrix}\right.\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Tacó:\left\{{}\begin{matrix}56x+65y=5,3\\x+y=0,25\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=1,2\\y=-0,97\end{matrix}\right.\left(vô\:lí\right)\)

Em xem lại đề nha!

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\\ a,\%m_{Fe}=\dfrac{0,02.56}{4,36}.100\approx25,688\%\\ \Rightarrow\%m_{Ag}\approx74,312\%\\ b,Ta.thấy:2,18=\dfrac{1}{2}.4,36\\ \Rightarrow m_{hh\left(câuB\right)}=\dfrac{1}{2}.m_{hh\left(câuA\right)}\\ n_{Fe}=\dfrac{0,02}{2}=0,01\left(mol\right)\\ n_{Ag}=\dfrac{2,18-0,01.56}{108}=0,015\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ 2Ag+Cl_2\rightarrow\left(t^o\right)2AgCl\\ n_{Cl_2}=\dfrac{3}{2}.n_{Fe}+\dfrac{1}{2}.n_{Ag}=\dfrac{3}{2}.0,01+\dfrac{1}{2}.0,015=0,0225\left(mol\right)\\ \Rightarrow V_{Cl_2\left(đktc\right)}=0,0225.22,4=0,504\left(l\right)\)

C32: 

a, \(n_C=\dfrac{2,4}{12}=0,2\left(mol\right)\)

PT: \(C+O_2\underrightarrow{t^o}CO_2\)

Theo PT: \(n_{CO_2}=n_C=0,2\left(mol\right)\Rightarrow V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

b, \(n_{NaOH}=0,3.1=0,3\left(mol\right)\)

\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=1,5\) → Pư tạo NaHCO₃ và Na₂CO₃

PT: \(CO_2+NaOH\rightarrow NaHCO_3\)

\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=n_{NaHCO_3}+n_{Na_2CO_3}=0,2\\n_{NaOH}=n_{NaHCO_3}+2n_{Na_2CO_3}=0,3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaHCO_3}=0,1\left(mol\right)\\n_{Na_2CO_3}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ m_NaHCO3 = 0,1.84 = 8,4 (g)

m_Na2CO3 = 0,1.106 = 10,6 (g)

c, \(C_{M_{NaHCO_3}}=C_{M_{Na_2CO_3}}=\dfrac{0,1}{0,3}=\dfrac{1}{3}\left(M\right)\)

Lần sau bạn đăng tách câu hỏi ra nhé.

C31:

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

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

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{14,6}.100\%\approx44,52\%\\\%m_{ZnO}\approx55,48\%\end{matrix}\right.\)

c, \(n_{ZnO}=\dfrac{14,6-0,1.65}{81}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Zn}+2n_{ZnO}=0,4\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,4.36,5}{10\%}=146\left(g\right)\)

a)

\(n_{HCl}=\dfrac{300.7,3\%}{36,5}=0,6\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            FeCO₃ + 2HCl --> FeCl₂ + CO₂ + H₂O

\(n_{khí}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PTHH, n_HCl(pư) = 2.n_khí = 0,2 (mol) < 0,6 (mol)

=> HCl dư

Gọi số mol Fe, FeCO₃ là a, b (mol)

=> \(\left\{{}\begin{matrix}56a+116b=8,6\\a+b=0,1\end{matrix}\right.\)

=> a = 0,05 (mol); b = 0,05 (mol)

=> \(\left\{{}\begin{matrix}m_{Fe}=0,05.56=2,8\left(g\right)\\m_{FeCO_3}=0,05.116=5,8\left(g\right)\end{matrix}\right.\)

n_FeCl2 = 0,1 (mol) => m_FeCl2 = 0,1.127 = 12,7 (g)

n_HCl(dư) = 0,6 - 0,2 = 0,4 (mol) => m_HCl(dư) = 0,4.36,5 = 14,6 (g)

m_{dd sau pư} = 8,6 + 300 - 0,05.2 - 0,05.44 = 306,3 (g)

\(\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{12,7}{306,3}.100\%=4,146\%\\C\%_{HCl\left(dư\right)}=\dfrac{14,6}{306,3}.100\%=4,767\%\end{matrix}\right.\)

b)

\(\overline{M}_X=\dfrac{0,05.2+0,05.44}{0,05+0,05}=23\left(g/mol\right)\)

=> \(d_{X/H_2}=\dfrac{23}{2}=11,5\)

a) PTHH: Fe + 2HCl ===> FeCl₂ + H₂

Fe₂O₃ + 6HCl ===> 2FeCl₃ + 3H₂O

Ta có: n_H2 = \(\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

=> n_Fe = 0,1(mol)

=> m_Fe = \(0,1\cdot56=5,6\left(gam\right)\)

=> %m_Fe = \(\dfrac{5,6}{8}\cdot100\%=70\%\)

=> %m_Fe2O3 = \(100\%-70\%=30\%\)

a,\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:     0,1     0,2         0,1            0,1

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

Mol:     0,05         0,3          0,1     

\(m_{Fe}=0,1.56=5,6\left(g\right);m_{Fe_2O_3}=13,6-5,6=8\left(g\right)\)

\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)

b,\(m_{HCl}=\left(0,2+0,3\right).98=49\left(g\right)\)

\(m_{ddHCl}=\dfrac{49.100}{10}=490\left(g\right)\)

c,\(m_{ddX}=13,6+490-0,1.2=503,4\left(g\right)\)

d,\(C\%_{FeCl_2}=\dfrac{0,1.127.100\%}{503,4}=2,52\%\)

  \(C\%_{FeCl_3}=\dfrac{0,05.162,5.100\%}{503,4}=1,614\%\)

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\left(1\right)\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ \Rightarrow n_{Fe}=n_{H_2}=n_{FeCl_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Fe}=\dfrac{0,1.56}{13,6}.100\approx41,176\%\\ \Rightarrow\%m_{Fe_2O_3}\approx58,824\%\\ b.n_{Fe_2O_3}=\dfrac{13,6-0,1.56}{160}=0,05\left(mol\right)\\ n_{HCl}=2.n_{Fe}+6.n_{Fe_2O_3}=2.0,1+6.0,05=0,5\left(mol\right)\\ m_{ddHCl}=\dfrac{0,5.36,5.100}{10}=182,5\left(g\right)\)

\(c.m_{ddX}=m_{hh\left(Fe,Fe_2O_3\right)}+m_{ddHCl}-m_{H_2}=13,6+182,5-0,1.2=195,9\left(g\right)\\ d.n_{FeCl_3}=2.0,05=0,1\left(mol\right)\\ C\%_{ddFeCl_3}=\dfrac{0,1.162,5}{195,9}.100\approx8,295\%\\ C\%_{ddFeCl_2}=\dfrac{0,1.127}{195,9}.100\approx6,483\%\)

\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\) 

        1        2               1           1

       0,1     0,2            0,1         0,1

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

         1            6              2             3

      0,05        0,3            0,1  

a) \(n_{Fe}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{Fe}=01,.56=5,6\left(g\right)\)

\(m_{Fe2O3}=13,6-5,6=8\left(g\right)\)

⁰/₀Fe =  \(\dfrac{5,6.100}{13,6}=41,18\)⁰/0

⁰/₀Fe₂O₃ = \(\dfrac{8.100}{13,6}=58,82\)⁰/₀

b) \(n_{HCl\left(tổng\right)}=0,2+0,3=0,5\left(mol\right)\)

⇒ \(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{10}=182,5\left(g\right)\)

c) \(n_{FeCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

⇒ \(m_{FeCl2}=0,1.127=12,7\left(g\right)\)

\(n_{FeCl3}=\dfrac{0,3.2}{6}=0,1\left(mol\right)\)

⇒ \(m_{FeCl3}=0,1.162,5=16,25\left(g\right)\)

d) \(m_{ddspu}=13,6+182,5-\left(0,1.2\right)=195,9\left(g\right)\)

\(C_{FeCl2}=\dfrac{12,7.100}{195,9}=6,48\)⁰/₀

\(C_{FeCl3}=\dfrac{16,25.100}{195,9}=8,3\)⁰/₀

 Chúc bạn học tốt

a) Bảo toàn nguyên tố H : \(n_{HCl}.1=2n_{H_2}=0,6\left(mol\right)\)

=> nH2=0,3(mol)

=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

b) Áp dụng định luật bảo toàn khối lượng :

 \(m_{ct}=m_{kl}+m_{HCl}-m_{H_2}=10,4+0,6.36,5-0,3.2=31,7\left(g\right)\)