a) $2R + 2nHCl \to 2RCl_n + nH_2$

Theo PTHH : 

$n_R = \dfrac{2}{n}.n_{H_2} = \dfrac{0,2}{n}(mol)$

$\Rightarrow \dfrac{0,2}{n}.R = 6,5 \Rightarrow R= \dfrac{65}{2}n$

Với n = 2 thì R = 65(Zn)

Vậy kim loại cần tìm là Zn

b)

$n_{HCl} = 2n_{H_2} = 0,2(mol)$
$m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)$

c)

Sau phản ứng : 

$m_{dd} = 6,5 + 100 - 0,1.2 = 106,3(gam)$

$C\%_{ZnCl_2} = \dfrac{0,1.136}{106,3}.100\% = 12,8\%$

a, Giả sử R có hóa trị n.

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

Ta có: \(n_R=\dfrac{4,8}{M_R}\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{n}{2}n_R=\dfrac{2,4n}{M_R}\left(mol\right)\)

Mà: m _{dd tăng} = m_R - m_H2 \(\Rightarrow4,4=4,8-\dfrac{2,4n}{M_R}.2\Rightarrow M_R=12n\left(g/mol\right)\)

Với n = 2, M_R = 24 (g/mol) là thỏa mãn.

Vậy: R là Mg.

b, Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{14,6}{10\%}=146\left(g\right)\)

Có: m _{dd sau pư} = 4,8 + 146 - 0,2.2 = 150,4 (g)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,2.136}{150,4}.100\%\approx18,085\%\)

Chọn A

Vì phản ứng không sinh ra chất khí hay kết tủa 

Vậy n = 2, R = 24 g/mol thỏa mãn, chất cần tìm là M g O H 2 .

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 \(HCl\)

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

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

PTHH: R + 2HCl ---> RCl₂ + H₂ (1)

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

\(n_{HCl}=\dfrac{100}{1000}.5=0,5\left(mol\right)\)

Ta thấy: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\)

Vậy HCl dư.

Theo PT₍₁₎: \(n_R=n_{H_2}=0,2\left(mol\right)\)

=> \(M_R=\dfrac{4,8}{0,2}=24\left(g\right)\)

Vậy R là magie (Mg)

PT: Mg + 2HCl ---> MgCl₂ + H₂ (2)

Ta có: \(m_{dd_{MgCl_2}}=4,8+\dfrac{100}{1000}-0,2.2=4,5\left(lít\right)\)

Theo PT₍₂₎: \(n_{MgCl_2}=n_{H_2}=0,2\left(mol\right)\)

=> \(C_{M_{MgCl_2}}=\dfrac{0,2}{4,5}=\dfrac{2}{45}M\)

PTHH: \(R+CuSO_4\rightarrow RSO_4+Cu\)

Ta có: \(n_{CuSO_4}=\dfrac{200\cdot16\%}{160}=0,2\left(mol\right)=n_{Cu}=n_R\)

\(\Rightarrow m_{Cu}=0,2\cdot64=12,8\left(g\right)\)

Theo bài ra, ta có: m_{hỗn hợp} = m_{R ban đầu} - m_{R pư} + m_Cu

\(\Rightarrow51,75-m_{R\left(pư\right)}+12,8=51,55\left(g\right)\)

\(\Rightarrow m_{R\left(pư\right)}=13\left(g\right)\) \(\Rightarrow M_R=\dfrac{13}{0,2}=65\) \(\Rightarrow\) R là Kẽm

Mặt khác: \(m_{dd}=m_{R\left(bđ\right)}+m_{ddCuSO_4}-m_{Cu}-m_{Zn\left(dư\right)}=225,95\left(g\right)\)

\(\Rightarrow C\%_{ZnSO_4}=\dfrac{0,2\cdot161}{225,95}\cdot100\%\approx14,25\%\)

PT: \(RO+2HCl\rightarrow RCl_2+H_2O\)

Gọi: m_RO = a (g)

\(\Rightarrow n_{RO}=\dfrac{a}{M_R+16}\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{HCl}=2n_{RO}=\dfrac{2a}{M_R+16}\left(mol\right)\\n_{RCl_2}=n_{RO}=\dfrac{a}{M_R+16}\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{HCl}=\dfrac{2a}{M_R+16}.36,5=\dfrac{73a}{M_R+16}\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{\dfrac{73a}{M_R+16}}{14,6\%}=\dfrac{500a}{M_R+16}\left(g\right)\)

⇒ m _{dd sau pư} = \(a+\dfrac{500a}{M_R+16}\left(g\right)\)

Mà: \(m_{RCl_2}=\dfrac{a}{M_R+16}.\left(M_R+71\right)\left(g\right)\)

\(\Rightarrow\dfrac{\dfrac{a}{M_R+16}.\left(M_R+71\right)}{a+\dfrac{500a}{M_R+16}}=0,17592\)

\(\Rightarrow M_R=24\left(g/mol\right)\)

→ R là Mg.

tham khảo

Coi mdd H2SO4=100(gam)mdd H2SO4=100(gam)
nR2(SO4)n=nR2(CO3)n=nH2SO4n=0,1n(mol)nR2(SO4)n=nR2(CO3)n=nH2SO4n=0,1n(mol)

Sau phản ứng , 

mR2(SO4)n=0,1n(2R+96n)=0,2Rn+9,6(gam)mR2(SO4)n=0,1n(2R+96n)=0,2Rn+9,6(gam)