\(300(ml)=0,3(l)\\ n_{HCl}=1.0,3=0,3(mol);n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ \text{LTL: }\dfrac{n_{Fe}}{1}<\dfrac{n_{HCl}}{2}\Rightarrow HCl\text{ dư}\\ \Rightarrow n_{HCl(dư)}=0,3-0,1.2=0,1(mol)\\ \Rightarrow m_{HCl(dư)}=0,1.36,5=3,65(g)\\ b,n_{FeCl_2}=n_{Fe}=0,1(mol)\\ \Rightarrow \begin{cases} C_{M_{FeCl_2}}=\dfrac{0,1}{0,3}=0,33M\\ C_{M_{HCl(dư)}}=\dfrac{0,1}{0,3}=0,33M \end{cases}\)

\(n_{HCl}=0,2.0,5=0,1(mol)\\ a,Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{Fe}=n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,05(mol)\\ b,m_{Fe}=0,05.56=2,8(g)\\ c,m_{FeCl_2}=0,05.127=6,35(g)\)

\(n_{Cu}=\dfrac{32}{64}=0,5\left(mol\right);n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ PTHH:2Cu+O_2\rightarrow2CuO\\ TL:....2.....1.....2\\ BR:....0,2......0,1.....0,2\left(mol\right)\)

Vì \(\dfrac{n_{Cu}}{2}>\dfrac{n_{O_2}}{1}\) nên sau p/ứ Cu dư, O₂ hết

\(a,m_{CuO}=0,2\cdot80=16\left(g\right)\)

\(m_{HCl}=\dfrac{200\cdot14,6\%}{100\%}=29,2\left(g\right)\\ \Rightarrow n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\\ PTHH:CuO+2HCl\rightarrow CuCl_2+H_2O\\ TL:....1.....2.....1.....1\\ BR:....0,2.....0,4.....0,2.....0,2\left(mol\right)\)

Vì \(\dfrac{n_{CuO}}{1}< \dfrac{n_{HCl}}{2}\) nên sau p/ứ HCl dư, CuO hết

\(\Rightarrow n_{HCl\left(dư\right)}=0,8-0,4=0,4\left(mol\right)\\ \Rightarrow m_{HCl\left(dư\right)}=14,6\left(g\right)\)

\(m_{CuCl_2}=0,2\cdot135=27\left(g\right)\\ m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\\ m_{CuO}=16\left(g\right)\\ m_{H_2O}=0,2\cdot18=3,6\left(g\right)\\ \Rightarrow m_{dd_{CuCl_2}}=16+14,6-3,6=27\left(g\right)\)

Vậy \(C\%_{dd_{CuCl_2}}=\dfrac{27}{200}\cdot100\%=13,5\%\)

Fe + 2HCl -> FeCl2 + H2
nFe = 5,6/56 = 0,1 mol
=>nH2 = 0,1 mol
=> VH2= 0,1*22,4= 2,24 lít

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

          0,1-->0,2------------------>0,1

=> \(\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2.36,5}{15\%}=\dfrac{146}{3}\left(g\right)\\V_{H_2}=0,1.22,4=4,48\left(l\right)\end{matrix}\right.\)

PTHH: \(Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_3+4H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\\n_{HCl}=\dfrac{300\cdot3,65\%}{36,5}=0,3\left(mol\right)\end{matrix}\right.\) 

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,3}{8}\) \(\Rightarrow\) Fe₃O₄ còn dư, HCl p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{Fe_3O_4\left(dư\right)}=0,0625\left(mol\right)\\n_{FeCl_2}=0,0375\left(mol\right)\\m_{FeCl_3}=0,075\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_3O_4\left(dư\right)}=0,0625\cdot232=14,5\left(g\right)\\m_{muối}=0,0375\cdot127+0,075\cdot162,5=16,95\left(g\right)\end{matrix}\right.\)

nFe3O4= 23,2/232=0,1(mol); nHCl = (300.3,65%)/36,5= 0,3(mol)

a) PTHH: Fe3O4 + 8 HCl -> 2 FeCl3 + FeCl2 + 4 H2O

b) Ta có: 0,3/8 < 0,1/1

=> Fe3O4 dư, HCl hết, tính theo nHCl.

=> nFe3O4(p.ứ)= nFeCl2= nHCl/8=0,3/8= 0,0375(mol)

=> mFe3O4(dư)= (0,1- 0,0375).232=14,5(g)

c) nFeCl3= 2/8. 0,3= 0,075(mol)

=> mFeCl3= 0,075.162,5=12,1875(g)

mFeCl2= 0,0375. 127=4,7625(g)

=>m(muối)= 12,1875+ 4,7625= 16,95(g)

\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 0,25.........0,5.........0,25.......0,25\left(mol\right)\\ a.V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\\ b.m_{HCl}=0,5.36,5=18,25\left(g\right)\\ c.n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\ Fe_2O_3+3H_2\underrightarrow{^{to}}2Fe+3H_2O\\ Vì:\dfrac{0,25}{3}< \dfrac{0,1}{1}\\ \Rightarrow Fe_2O_3dư\\ n_{Fe}=\dfrac{2}{3}.0,25=\dfrac{1}{6}\left(mol\right)\\ \Rightarrow m_{Fe}=\dfrac{1}{6}.56\approx9,333\left(g\right)\)

a,\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:    0,25     0,5                      0,25

\(\Rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)\)

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

c,\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ → 2Fe + 3H₂O

Mol:                   0,25      \(\dfrac{1}{6}\)

Ta có: \(\dfrac{0,1}{1}>\dfrac{0,25}{3}\)⇒ Fe₂O₃ dư, H₂ hết

\(m_{Fe}=\dfrac{1}{6}.56=9,33\left(g\right)\)

\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)

Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

          1        2              1            1

        0,25    0,5                        0,25

a) \(n_{H2}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)

\(V_{H2\left(dktc\right)}0,25.22,4=5,6\left(l\right)\)

b) \(n_{HCl}=\dfrac{0,25.2}{1}=0,5\left(mol\right)\)

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

c) \(n_{Fe2O3}=\dfrac{16}{160}=0,1\left(mol\right)\)

Pt : \(3H_2+Fe_2O_3\rightarrow\left(t_o\right)2Fe+3H_2O|\)

         3           1                  2           3

       0,25       0,1                \(\dfrac{1}{6}\)

Lập tỉ số so sánh : \(\dfrac{0,25}{3}< \dfrac{0,1}{1}\)

                  ⇒ H₂ phản ứng hết , Fe₂O₃ dư

                  ⇒ Tính toán dựa vào số mol của H₂

\(n_{Fe}=\dfrac{0,25.2}{3}=\dfrac{1}{6}\left(mol\right)\)

⇒ \(m_{Fe}=\dfrac{1}{6}.56=9,3\left(g\right)\)

 Chúc bạn học tốt

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ b,n_{HCl}=2n_{H_2}=0,3(mol);n_{FeCl_2}=n_{H_2}=0,15(mol)\\ \Rightarrow m_{HCl}=0,3.36,5=10,95(g)\\ m_{FeCl_2}=0,15.127=19,05(g)\)