a)\(n_{HCl}=0,5\cdot1=0,5mol\)

PT: \(2R+2nHCl\rightarrow2RCl_n+nH_2\)

\(n_R=\dfrac{4,05}{M_R}\)

\(n_{H_2}=\dfrac{5,04}{22,4}=0,225mol\)

\(\Rightarrow\dfrac{4,05}{M_R}\cdot n=0,225\cdot2\)

R là kim loại: 

\(\begin{matrix}n&1&2&3\\M_R&9&18&27\end{matrix}\)

Vậy R có hóa trị III và \(M_R=27\left(Al\right)\)

b)PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

          0,15    0,5                          0,225

          0,15    0,45        0,15         0,225

Vậy \(C_{M_{AlCl_3}}=\dfrac{0,15}{0,5}=0,3M\)

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

n_HCl = 0,05.2 = 0,1

Có 2.n_H2 < n_HCl => R phản ứng hết

PTHH: 2R + 6HCl --> 2RCl₃ + 3H₂

____0,02<-----------------------0,03

=> \(M_R=\dfrac{0,54}{0,02}=27\left(Al\right)\)

b) 

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

___________0,06<----0,02<---0,03

=> \(\left\{{}\begin{matrix}C_{M\left(HCldư\right)}=\dfrac{0,1-0,06}{0,05}=0,8M\\C_{M\left(AlCl_3\right)}=\dfrac{0,02}{0,05}=0,4M\end{matrix}\right.\)

a, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PT: \(M+H_2SO_4\rightarrow MSO_4+H_2\)

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

\(\Rightarrow M_M=\dfrac{11,2}{0,2}=56\left(g/mol\right)\)

→ M là Fe.

b, Theo PT: \(n_{FeSO_4}=n_{H_2SO_4\left(pư\right)}=n_{H_2}=0,2\left(mol\right)\)

⇒ n_{H2SO4 dư} = 0,5.1 - 0,2 = 0,3 (mol)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\\C_{M_{FeSO_4}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\end{matrix}\right.\)

c, Ta có: \(n_{FeSO_4.7H_2O}=n_{FeSO_4}=0,2\left(mol\right)\)

\(\Rightarrow m_{FeSO_4.7H_2O}=0,2.278=55,6\left(g\right)\)

câu a

dạ em làm xong câu B rồi mọi người khỏi cần trả lời nữa ạ

mik sửa lại cái dưới bị lỗi latex

\(a.n_{HCl}=0,05.2=0,1\left(mol\right);n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\ 2R+6HCl\rightarrow2RCl_3+3H_2\\ \Rightarrow\dfrac{0,1}{6}>\dfrac{0,03}{3}\Rightarrow HCl.dư,R.pư.hết\\ n_R=0,03.2:3=0,02\left(mol\right)\\ M_R=\dfrac{0,54}{0,02}=27\left(g/mol\right)\\ \Rightarrow R=27\left(Al,nhôm\right)\\ b.C_{M_{AlCl_3}}=\dfrac{0,3.2:3}{0,05}=0,4M\\ C_{M_{HCl\left(dư\right)}}=\dfrac{0,1-\left(0,3.6:3\right)}{0,05}=0,8M\)

\(a.n_{HCl}=0,05.2=0,1\left(mol\right)\\ n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\ 2R+6HCl\rightarrow2RCl_3+3H_2\\ \Rightarrow\dfrac{0,1}{6}>\dfrac{0,03}{3}\Rightarrow HCl.dư,R.pư.hết\\ n_R=0,03.2:3=0,02\left(mol\right)\\ M_R=\dfrac{0,54}{0,02}=27\left(g/mol\right)\\ \Rightarrow R=27\left(Al,nhôm\right)\\ b.n_{AlCl_3}=n_{Al}=0,02mol\\ C_{M_{AlCl_3}}=\dfrac{0,02}{0,05}=0,4M\\ C_M_{HCl\left(dư\right)}=\dfrac{0,1-\left(0,03.2\right)}{0,05}=0,8M\)

          \(2R+6HCl\rightarrow2RCl_3+3H_2\)

TPT:   2         6              2            3      (mol)

TĐB:  0,02   0,1           0,02      0,03    (mol)

PƯ:    0,02   0,06         0,02      0,03    (mol)

Dư:      0       0,04          0             0      (mol)

        50ml = 0,05 lít

\(n_{HCl}=C_M.V_{dd}=2.0,05=0,1\left(mol\right)\)

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

Tỉ lệ: \(\dfrac{n_{HCl}}{6}=\dfrac{0,1}{6}>\dfrac{n_{H_2}}{3}=\dfrac{0,03}{3}\)\(\Rightarrow n_{HCl}\) dư

      \(m_R=n.M\)

\(\Leftrightarrow0,54=0,02R\)

\(\Leftrightarrow R=27\)

Vậy kim loại R là Al (III)

       \(RCl_3\) là \(AlCl_3\)

Sau phản ứng còn \(AlCl_3\) và 0,04 mol \(HCl\)

\(C_{M_{HCl\left(dư\right)}}=\dfrac{n}{V_{dd}}=\dfrac{0,04}{0,05}=0,8\left(M\right)\)

\(C_{M_{AlCl_3}}=\dfrac{n}{V_{dd}}=\dfrac{0,02}{0,05}=0,4\left(M\right)\)

\(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

  \(M+2HCl\rightarrow MCl_2+H_2\)

0,15<-0,3<---0,15<----0,15

a. \(M=\dfrac{8,4}{0,15}=56\left(g/mol\right)\)

Vậy M là kim loại Fe.

b. \(n_{NaOH}=0,5.1=0,5\left(mol\right)\)

\(HCl+NaOH\rightarrow NaCl+H_2O\)

0,2<-----0,2

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)

0,15----->0,3

\(m_{dd.HCl}=\dfrac{\left(0,3+0,2\right).36,5.100\%}{10\%}=182,5\left(g\right)\)

\(m_{dd.A}=8,4+182,5-0,15.2=190,6\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{127.0,2.100\%}{190,6}=13,33\%\)

\(C\%_{HCl.dư}=\dfrac{0,3.36,5.100\%}{190,6}=5,75\%\)

PTHH: \(2M+2xHCl\rightarrow2MCl_x+xH_2\uparrow\)  (x là hóa trị của M)

Tính theo sản phẩm

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\) 

\(\Rightarrow n_M=\dfrac{0,4}{x}\left(mol\right)\) \(\Rightarrow M=\dfrac{4,8}{\dfrac{0,4}{x}}=12x\)

Ta thấy với \(x=2\) thì \(M=24\)  (Magie)

Mặt khác: \(n_{HCl}=\dfrac{50\cdot36,5\%}{36,5}=0,5\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{MgCl_2}=n_{H_2}=0,2\left(mol\right)\\n_{HCl\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\)

Ta lại có: \(m_{dd\left(sau.p/ứ\right)}=m_{Mg}+m_{ddHCl}-m_{H_2}=4,8+50-0,2\cdot2=54,4\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,2\cdot95}{54,4}\cdot100\%\approx34,93\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1\cdot36,5}{54,4}\cdot100\%\approx6,71\%\end{matrix}\right.\)