Đáp án C

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

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

Gọi kim loại cần tìm là A

a) PTHH: \(A+H_2O\rightarrow AOH+\dfrac{1}{2}H_2\uparrow\)

                 \(AOH+HCl\rightarrow ACl+H_2O\)

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

\(\Rightarrow M_A=\dfrac{7,8}{0,2}=39\) \(\Rightarrow\) Kim loại cần tìm là Kali

b) Ta có: \(\left\{{}\begin{matrix}n_{KCl}=0,2mol\\n_{HCl\left(pư\right)}=0,2mol\Rightarrow n_{HCl\left(dư\right)}=0,2\cdot20\%=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{KCl}=0,2\cdot74,5=14,9\left(g\right)\\m_{HCl\left(dư\right)}=0,04\cdot36,5=1,46\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{H_2}=2\cdot0,1=0,2\left(g\right)\)

\(\Rightarrow m_{dd}=m_K+m_{ddHCl}-m_{H_2}=7,8+\dfrac{0,24\cdot36,5}{10\%}-0,2=95,2\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{KCl}=\dfrac{14,9}{95,2}\cdot100\%\approx15,65\%\\C\%_{HCl\left(dư\right)}=\dfrac{1,46}{95,2}\cdot100\%\approx1,53\%\end{matrix}\right.\)

\(nAl=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(mHCl=\dfrac{200.7,3\%}{100\%}=14,6\left(g\right)\)

\(nHCl=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

PTHH:

\(2Al+6HCl\rightarrow2AlCl_2+3H_2\)

 2         6              2            3    (mol)

0,1       0,3         0,1           0,15   (mol)

LTL : 0,1 / 2 < 0,4/6

=> Al đủ , HCl dư

1. \(VH_2=0,15.22,4=3,36\left(l\right)\)

2. \(mH_2=0,15.2=0,3\left(g\right)\)

mdd = mAl + mddHCl - mH2 = 2,7 + 200 - 0,3 = 202,4 (g)

\(mH_2SO_{4\left(dưsaupứ\right)}=0,1.98=9,8\left(g\right)\)

\(mAlCl_2=0,1.98=9,8\left(g\right)\)

\(C\%_{ddH_2SO_4}=\dfrac{9,8.100}{202,4}=4,84\%\)

\(C\%_{AlCl_2}=\dfrac{9,8.100}{202,4}=4,84\%\)

câu a

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

a) 

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,2-->0,4----->0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

b) m_HCl = 0,4.36,5 = 14,6 (g)

=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)

c)

m_{dd sau pư} = 13 + 200 - 0,2.2 = 212,6 (g)

m_ZnCl2 = 0,2.136 = 27,2 (g)

=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)

\(CT:ACO_3\)

\(n_{CO_2}=\dfrac{4.4912}{22.4}=0.2005\left(mol\right)\)

\(ACO_3+2HCl\rightarrow ACl_2+CO_2+H_2O\)

\(0.2005.....0.401.....0.2005...0.2005\)

\(M_{ACO_3}=\dfrac{20.05}{0.2005}=100\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow A=100-60=40\)

\(A:Ca\left(Canxi\right)\)

\(m_{CaCl_2}=0.2005\cdot111=22.2555\left(g\right)\)

\(m_{dd}=20.05+100-0.2005\cdot44=111.228\left(g\right)\)

\(C\%_{CaCl_2}=\dfrac{22.2555}{111.228}\cdot100\%=20\%\)