Quy hết kim loại R (II) và Fe về M (II):

\(n_{HCl}=0,2.2=0,4\left(mol\right)\)

PTHH: M + 2HCl ---> MCl₂ + H₂↑

           0,2<-0,4------->0,2---->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

\(b,M_M=\dfrac{8}{0,2}=40\left(\dfrac{g}{mol}\right)\\ \rightarrow M_R=2.40-56=24\left(\dfrac{g}{mol}\right)\)

=> R là Mg

c, Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\)

PTHH:

Fe + 2HCl ---> FeCl₂ + H₂

a---->2a--------->a

Mg + 2HCl ---> MgCl₂ + H₂

b---->2b-------->b

=> hệ pt \(\left\{{}\begin{matrix}56a+24b=8\\2a+2b=0,4\end{matrix}\right.\Leftrightarrow a=b=0,1\left(mol\right)\left(TM\right)\)

=> \(\left\{{}\begin{matrix}m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Mg}=0,1.24=2,4\left(g\right)\end{matrix}\right.\)

d, \(\left\{{}\begin{matrix}C_{M\left(FeCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\\C_{M\left(MgCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\end{matrix}\right.\)

Gọi kim loại cần tìm là R (II)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_R=b\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow56a+M_R.b=8\left(g\right)\)

\(n_{HCl}=0,2.2=0,4\left(mol\right)\)

PTHH:

Fe + 2HCl --->FeCl₂ + H₂

a---->2a

R + 2HCl ---> RCl₂ + H₂

b---->2b

Theo pthh: \(n_{hhkl}=n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, Lập được hệ pt: \(\left\{{}\begin{matrix}56a+M_R.b=8\\2a+2b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,1\left(mol\right)\\M_R=24\left(\dfrac{g}{mol}\right)\end{matrix}\right.\)

=> R là Mg

c, \(\left\{{}\begin{matrix}n_{Fe}=0,1\left(mol\right)\\n_{Mg}=0,1\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Mg}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)

a) Gọi số mol Fe, Zn là a, b (mol)

=> 56a + 65b = 12,1 (1)

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

              a-->2a-------->a---->a

            Zn + 2HCl --> ZnCl₂ + H₂

              b---->2b---->b------->b

=> a + b = 0,2 (2)

(1)(2) => a = 0,1 (mol); b = 0,1 (mol)

\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{12,1}.100\%=46,28\%\\\%m_{Zn}=\dfrac{0,1.65}{12,1}.100\%=53,72\%\end{matrix}\right.\)

b) n_HCl = 2a + 2b = 0,4 (mol)

=> \(V_{dd.HCl}=\dfrac{0,4}{1}=0,4\left(l\right)\)

c) \(\left\{{}\begin{matrix}C_{M\left(FeCl_2\right)}=\dfrac{0,1}{0,4}=0,25M\\C_{M\left(ZnCl_2\right)}=\dfrac{0,1}{0,4}=0,25M\end{matrix}\right.\)

giúp e vs huhu

a) \(\left(2m-1\right)sinx+1-m=0\Rightarrow sinx=\dfrac{m-1}{2m-1}\)

     Pt có nghiệm:  \(-1\le\dfrac{m-1}{2m-1}\le1\)

                           \(\Rightarrow1-2m\le m-1\le2m-1\Rightarrow m\ge\dfrac{2}{3}\)

b) \(\left(m+1\right)sin3x-cos3x=m+2\)

    Pt có nghiệm:   \(\left(m+1\right)^2+\left(-1\right)^2\ge\left(m+2\right)^2\)

                           \(\Rightarrow m^2+2m+1+1\ge m^2+4m+4\)

                           \(\Rightarrow-2m\ge2\Rightarrow m\le-1\)

a, \(\left(2m-1\right)sinx+1-m=0\)

\(\Leftrightarrow sinx=\dfrac{m-1}{2m-1}\)

Phương trình có nghiệm khi:

\(-1\le\dfrac{m-1}{2m-1}\le1\)

\(\Leftrightarrow\left[{}\begin{matrix}m>\dfrac{2}{3}\\m\le0\end{matrix}\right.\)