nH2 = 0,336/22,4 = 0,015 (mol)

PTHH: 2A + 6HCl -> 2ACl3 + 3H2

nACl3 = nA = 0,015 : 3 . 2 = 0,01 (mol)

M(A) = 0,27/0,01 = 27 (g/mol)

=> A là Al

mAlCl3 = 0,01 . 133,5 = 1,335 (g)

Câu 6.

\(n_{H_2}=\dfrac{0,336}{22,4}=0,015mol\)

\(2A+6HCl\rightarrow2ACl_3+3H_2\)

0,01                                0,015

\(\overline{M_A}=\dfrac{0,27}{0,01}=27đvC\)

\(\Rightarrow A\) là Al nhôm.

\(m_{AlCl_3}=0,01\cdot133,5=1,335g\)

1.a) \(A+HCl\rightarrow ACl+\dfrac{1}{2}H_2\)

Ta có : \(n_A=2n_{H_2}=2.\dfrac{2,8}{22,4}=0,25\left(mol\right)\)

\(\Rightarrow M_A=\dfrac{9,75}{0,25}=39\)

Vậy A là Kali (K)

b)\(n_{HCl}=2n_{H_2}=0,25\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,25.26,5=9,125\left(g\right)\)

\(n_{H_2}=\dfrac{1,85925}{24,79}=0,075\left(mol\right)\)

\(m_{dd.HCl}=500.1,2=600\left(g\right)\)

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

0,05<-0,15<--0,05<----0,075

a. \(R=\dfrac{1,35}{0,05}=27\left(g/mol\right)\)

Vậy tên kim loại là nhôm (Al)

b. 

\(n_{AgCl}=\dfrac{14,35}{143,5}=0,1\left(mol\right)\)

\(HCl+AgNO_3\rightarrow AgCl+HNO_3\)

0,025<-------------0,025

\(AlCl_3+3Ag\left(NO_3\right)\rightarrow3AgCl+Al\left(NO_3\right)_3\)

0,025------------------->0,075

\(CM_{HCl.đã.dùng}=\dfrac{0,025}{0,5}=0,05M\)

c.

\(m_{dd.X}=1,35+600-0,075.2=601,2\left(g\right)\)

\(n_{HCl.dư}=0,025.2=0,05\left(mol\right)\)

\(C\%_{AlCl_3}=\dfrac{0,05.133,5.100\%}{601,2}=1,11\%\)

\(C\%_{HCl.dư}=\dfrac{0,05.36,5.100\%}{601,2}=0,3\%\)

ta có mBaCl2=49,92g=>nBaCl2=0,24mol

n AClx=0.2*0.8=0.16mol

pt: xBaCl2 + A2(SO4)x->xBaSO4 + 2 AClx

mol 0.24 0.08 0.16

=>0.24*2=0.16x=>x=3=> A có hóa tri 3

=>M A2(SO4)3=27.36/0.08=342=>A=27

=>A là nhôm (Al) và công thức của muối là Al2(SO4)3

1/ nNaCl=5,85/58,5=0,1 mol. 
nAgNO3=34/170=0,2 mol. 
PTPU: NaCl+AgNO3=>AgCl+NaNO3 
vì NaCl và AgNO3 phan ung theo ti le 1:1 (nAgNO3 p.u=nNaCl=0,1 mol) 
=>AgNO3 du 
nAgNO3 du= 0,2-0,1=0,1 mol. 
Ta tinh luong san pham theo chat p.u het la NaCl 
sau p.u co: AgNO3 du:0,1 mol; AgCl ket tua va NaCl: nAgCl=nNaNO3=nNaCl=0,1 mol.V(dd)=300+200=500ml=0,5 ()l 
=>khoi lg ket tua: mAgCl=0,1.143,5=14,35 g 
C(M)AgNO3=C(M)NaNO3=n/V=0,1/0,5=0,2 M

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+3H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ a,n_{HCl}=2.n_{H_2}=2.0,4=0,8\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,8}{8}=0,1\left(l\right)\\ b,FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\downarrow\\ AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ n_{AgCl}=n_{AgNO_3}=3.n_{AlCl_3}+2.n_{FeCl_2}=3.a+2.b=3.0,2+2.0,1=0,8\left(mol\right)\\ \Rightarrow a=\dfrac{170.0,8}{250}.100=54,4\%\\ b=m_{\downarrow}=m_{AgCl}=0,8.143,5=114,8\left(g\right)\)