a. Gọi n là hóa trị của kim loại R

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

TĐB: \(\dfrac{4,05}{R}\) - \(\dfrac{4,05n}{R}\) - \(\dfrac{4,05}{R}\) -  \(\dfrac{2,025n}{R}\)    (mol)

\(n_R=\dfrac{m}{M}=\dfrac{4,05}{R}\left(mol\right)\)

\(m_{H_2}=n.M=\dfrac{2,025n}{R}.2=\dfrac{4,05n}{R}\left(g\right)\)

\(m_{ddRCl_n}=m_R+m_{ddHCl}-m_{H_2}\)

\(116,1=4,05+112,5-\dfrac{4,05n}{R}\)

\(\dfrac{4,05n}{R}=116,55-116,1\)

\(\dfrac{4,05n}{R}=0,45\)

\(4,05n=0,45R\)

\(\dfrac{4,05}{0,45}=\dfrac{R}{n}\)

\(9=\dfrac{R}{n}\)

\(9n=R\)

Nếu \(n=1\Rightarrow R=9\) (loại)

       \(n=2\Rightarrow R=18\) (loại)

       \(n=3\Rightarrow R=27\) (chọn)

Vậy kim loại R là Al

b) Kim loại tìm được là Al (III)

          \(2Al+6H_2SO_{4\left(đ,t^0\right)}\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)

TĐB: \(0,15\) - \(0,45\)                                                          (mol)

\(n_{Al}=\dfrac{m}{M}=\dfrac{4,05}{27}=0,15\left(mol\right)\)

\(m_{H_2SO_4}=n.M=0,45.98=44,1\left(g\right)\)

m\(m_{ddH_2SO_4}=\dfrac{m_{ct}.100\%}{C\%}=\dfrac{44,1.100\%}{98\%}=45\left(g\right)\)

\(n_{SO_3}=\dfrac{30}{80}=0,375\left(mol\right)\)

PTHH: SO₃ + H₂O ---> H₂SO₄

        0,375 ----------------> 0,375

\(C_{M\left(H_2SO_4\right)}=\dfrac{0,375}{0,15}=2,5M\)

PTHH: 2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

           0,25 <- 0,375

\(m_{Al}=0,25.27=6,75\left(g\right)\)

\(Fe+H_2SO_4 \to FeSO_4+H_2\\ n_{H_2}=0,15(mol)\\ a/\\ n_{Fe}=n_{H_2}=0,15(mol)\\ m_{Fe}=0,15.56=8,4(g)\\ b/\\ n_{H_2SO_4}=n_{H_2}=0,15(mol)\\ CM_{H_2SO_4}=\dfrac{0,15}{2}=0,75M c/\\ n_{FeSO_4}=n_{H_2}=0,15(mol)\\ CM_{FeSO_4}=\dfrac{0,15}{0,2}=0,75M\\\)

nHCl=0,3.2=0,6(mol)

a) PTHH: CuO +2 HCl -> CuCl2 + H2O

0,3_______________0,6___0,3(mol)

b) mCuO=0,3.80=24(g)

c) VddCuCl2=VddHCl=0,3(l)

=>CMddCuCl2=0,3/0,3=1(M)

d) m(muối)=0,3.135=40,5(g)

Bài 9 : 

\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

0,05--->0,1-------->0,05 

a) \(C_{MddHCl}=\dfrac{0,1}{0,1}=1\left(M\right)\)

b) \(m_{CuCl2}=0,05.135=6,75\left(g\right)\)

c) \(C_{MCuCl2}=\dfrac{0,05}{0,1}0,5\left(M\right)\)

Câu 10 : 

\(n_{FeO}=\dfrac{3,6}{72}=0,05\left(mol\right)\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

0,05-->0,1------->0,05

\(m_{ddHCl}=\dfrac{0,1.36,5}{10\%}100\%=36,5\left(g\right)\)

\(m_{ddspu}=3,6+36,5=40,1\left(g\right)\)

\(C\%_{FeCl2}=\dfrac{0,05.127}{40,1}.100\%=15,84\%\)

Câu 10 :

\(n_{FeO}=\dfrac{3,6}{72}=0,05\left(mol\right)\)

PTHH :

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

0,05      0,1           0,05       0,05

\(m_{ddHCl}=\dfrac{0,1.36,5.100}{10}=36,5\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{0,05.127}{\left(3,6+36,5\right)}.100\%=15,84\%\)

\(a,PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4}=n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\\ \Rightarrow m_{CuSO_4}=0,1\cdot160=16\left(g\right)\\ b,n_{H_2SO_4}=n_{CuO}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{H_2SO_4}}=0,1\cdot98=9,8\left(g\right)\\ \Rightarrow C\%_{H_2SO_4}=\dfrac{9,8}{200}\cdot100\%=4,9\%\)