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![](https://rs.olm.vn/images/avt/0.png?1311)
a,b)R + 2HCl---->RCl2+H2(1).
4,8/MR-------------------------4,8/MR.
vì nH2=0,2=>4,8/MR=0,2=>MR=24=>R là Magie (Mg)
c)m muối=mkim loại +mCl-=4,8+35,5.0,4=19 gam.
a,b)R + 2HCl---->RCl2+H2(1).
4,8/MR-------------------------4,8/MR.
vì nH2=0,2=>4,8/MR=0,2=>MR=24=>R là Magie (Mg)
c) khối lượng muối =mMgCl2=0.2.MMgCl2=19 gam.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: R + 2HCl --> RCl2 + H2
_____0,25<----------------0,25
=> \(\dfrac{34,25}{0,25}=137\left(Ba\right)\)
b) Muối thu được là BaCl2
Hiệu độ âm điện = 3,16 - 0,89 = 2,27
=> Liên kết ion
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(R+2HCl\rightarrow RCl_2+H_2\)
\(0.1........0.2................0.1\)
\(M_R=\dfrac{13.7}{0.1}=137\left(\dfrac{g}{mol}\right)\)
\(R:Ba\)
\(200\left(ml\right)=0.2\left(l\right)\)
\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
b)
PTHH: 2R + 2H2O --> 2ROH + H2
_____0,2<--------------0,2<----0,1
=> \(M_R=\dfrac{7,8}{0,2}=39\left(K\right)\)
c)
\(C_{M\left(KOH\right)}=\dfrac{0,2}{0,5}=0,4M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,n_{CO_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ PTHH:M_2CO_3+2HCl\to 2MCl+H_2O+CO_2\uparrow\\ \Rightarrow n_{M_2CO_3}=n_{CO_2}=0,15(mol)\\ \Rightarrow M_{M_2CO_3}=\dfrac{15,9}{0,15}=106(g/mol)\\ \Rightarrow M_{M}=\dfrac{106-12-16.3}{2}=23(g/mol)\)
Vậy M là natri (Na)
\(b,n_{HCl}=2n_{CO_2}=0,3(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,3}{0,75}=0,4(l)\\ X:NaCl\\ n_{NaCl}=n_{HCl}=0,3(mol)\\ \Rightarrow C_{M_{NaCl}}=\dfrac{0,3}{0,4}=0,75M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(PTHH:2R+2nHCl\rightarrow2RCl_n+nH_2\)
\(\left(mol\right)\) \(\dfrac{0,72}{R}\) \(\dfrac{0,72n}{R}\) \(\dfrac{0,72}{R}\) \(\dfrac{0,36n}{R}\)
\(a.\)Theo đề bài ta có:
\(0,72+120-\dfrac{0,36n}{R}.2=120,66\Leftrightarrow R=12n\)
Biện luận: \(\Rightarrow\left\{{}\begin{matrix}n=2\\R=24\left(Mg:Magie\right)\end{matrix}\right.\)
\(b.\) Khi đó: \(n_{HCl}=0,06\left(mol\right)\rightarrow a=\dfrac{0,06.36,5}{120}.100\%=1.825\left(\%\right)\)
\(c.C\%_{ddMgCl_2}=\dfrac{0,03.95}{120,66}.100\%=2,362\left(\%\right)\)
$m_{H_2} = 0,72 + 120 -120,66 = 0,06(gam)$
$n_{H_2} = 0,03(mol)$
Gọi n là hóa trị của R
$2R + 2nHCl \to 2RCl_n + nH_2$
Theo PTHH :
$n_R = \dfrac{2}{n}n_{H_2} = \dfrac{0,06}{n}(mol)$
Suy ra: $\dfrac{0,06}{n}.R = 0,72 \Rightarrow R = 12n$
Với n = 2 thì R = 24(Magie)
b)
$n_{HCl} = 2n_{H_2} = 0,06(mol)$
$C\%_{HCl} = \dfrac{0,06.36,5}{120}.100\% = 1,825\%$
(a = 1,825)
c)
$C\%_{MgCl_2} = \dfrac{0,03.95}{120,66}.100\% = 2,36\%$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ KL:M\\ M+2HCl\rightarrow MCl_2+H_2\\ n_{MCl_2}=n_M=n_{H_2}=0,05\left(mol\right)\\ M_{MCl_2}=\dfrac{4,75}{0,05}=95\left(\dfrac{g}{mol}\right)\\ M\text{à}:M_{MCl_2}=M_M+71\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_M+71=95\\ \Leftrightarrow M_M=24\left(\dfrac{g}{mol}\right)\\ \Rightarrow M\left(II\right):Magie\left(Mg=24\right)\\ a=24.0,05=1,2\left(g\right)\)