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a)Gọi kim loại cần tìm là R
nAg =0,7(mol)
\(R+2AgNO_3\rightarrow R\left(NO_3\right)_2+2Ag\)
nR =\(\dfrac{1}{2}\)nAg =0,35(mol)
MR=\(\dfrac{19,6}{0,35}\)=56
kim loại là Fe
b)nAgNO3=nAg=0,7
CMAgNO3=\(\dfrac{0,7.1000}{140}\)=5
c)nFe(NO3)=0,35
mdd sau pứ=19,6+140=159,6
CmFe(NO3)=\(\dfrac{0,35.1000}{159,6}=2,19\)
a,R +2 AgNO3--> R(NO3)2 + 2Ag
Ta có nAg=75,6/108=0,945mol
Theo PTHH ta có: nR=nAg/2
<=> 19,6/R=0,945/2
<=> R=41 g/mol=40g /mol
vậy R là Ca
b, Ta có nAgNO3=nAg=0,945mol
=> CM dd AgNO3=0,945/0,14=6,75M
c, Ta có nCa(NO3)2=nAg/2=0,475mol
=> CM dd Ca(NO3)2=0,475/0,14=3,39M
\(m_{dd.HCl}=1,08.150=162\left(g\right)\)
\(n_{CO_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
\(RCO_3+2HCl\rightarrow RCl_2+H_2O+CO_2\)
0,15<----------------0,15<-----------0,15
Có: \(R+60=\dfrac{12,6}{0,15}\Rightarrow R=24\left(g/mol\right)\)
a. Kim loại R là Magie (Mg)
b. \(C\%_{MgCl_2}=\dfrac{0,15.95.100\%}{12,6+162-0,15.44}=8,48\%\)
c. \(n_{AgCl}=\dfrac{53,8125}{143,5}=0,375\left(mol\right)\)
\(MgCl_2+2AgNO_3\rightarrow2AgCl+Mg\left(NO_3\right)_2\)
0,15-------------------->0,3
Vì \(n_{AgCl}=0,3\left(mol\right)< 0,375\left(mol\right)_{theo.đề}\) \(\Rightarrow\) HCl dư
\(HCl+AgNO_3\rightarrow AgCl+HNO_3\)
0,075<------------0,075
\(CM_{HCl.đem.dùng}=\dfrac{0,075}{0,15}=0,5M\)
n(AgNO3) = 1.0,2 = 0,2mol
2AgNO3 + M → M(NO3)2 + 2Ag
0,2 0,1 m
[M(NO3)2] = 0,1(M+124) = 18,8g
→ M = 64. M là Cu
=> Đáp án D
\(n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(2A+2nH_2O\rightarrow2A\left(OH\right)_n+nH_2\)
\(\dfrac{0.1}{n}........................0.05\)
\(M_A=\dfrac{3.9}{\dfrac{0.1}{n}}=39n\)
Với : \(n=1\rightarrow A=39\)
\(A:K\)
\(m_{KOH}=0.1\cdot56=5.6\left(g\right)\)
\(m_{ddX}=3.9+46.2-0.05\cdot2=50\left(g\right)\)
\(C\%_{KOH}=\dfrac{5.6}{50}\cdot100\%=11.2\%\)
\(b.\)
\(K_2O+H_2O\rightarrow2KOH\)
\(0.1....................0.2\)
\(m_{KOH}=0.2\cdot56=11.2\left(g\right)\)
\(m_{dd_X}=\dfrac{11.2}{28\%\%}=40\left(g\right)\)
Bài 6:
\(R_2O+H_2O\rightarrow2ROH\\ n_{ROH}=0,2\left(mol\right)\rightarrow n_{R_2O}=0,1\left(mol\right)\\ \rightarrow M_{R_2O}=\dfrac{6,2}{0,1}=62\left(\dfrac{g}{mol}\right)\\Lại.có:M_{R_2O}=2M_R+16\\ \Rightarrow2M_R+16=62\\ \Leftrightarrow M_R=23\left(\dfrac{g}{mol}\right)\\ \Rightarrow R\left(I\right):Natri\left(Na=23\right)\)
Bài 7:
\(n_{H_2}=0,4\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \rightarrow\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{11}.100\approx49,091\%\\\%m_{Fe}\approx50,909\%\end{matrix}\right.\)
Gọi kim loại hóa trị II cần tìm là A.
\(A+Cl_2\underrightarrow{to}ACl_2\\ ACl_2+2AgNO_3\rightarrow A\left(NO_3\right)_2+2AgCl\\ m_{\downarrow}=m_{AgCl}=86,1\left(g\right)\\ n_{AgCl}=\dfrac{86,1}{143,5}.100=0,6\left(mol\right)\\ n_A=n_{ACl_2}=n_{Cl_2}=n_{A\left(NO_3\right)_2}=\dfrac{0,6}{2}=0,3\left(mol\right)\\ M_A=\dfrac{41,1}{0,3}=137\left(\dfrac{g}{mol}\right)\\ \rightarrow A:Bari\left(Ba=137\right)\\ b.V_{Cl_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c.m_{muối}=m_{Ba\left(NO_3\right)_2}=0,3.261=78,3\left(g\right)\)
\(a.A+2HCl\rightarrow ACl_2+H_2\\ ACl_2+2AgNO_3\rightarrow A\left(NO_3\right)_2+2AgCl\downarrow\\ n_{AgCl\downarrow}=\dfrac{57,4}{143,5}=0,4\left(mol\right)\\ n_A=n_{ACl_2}=\dfrac{n_{AgCl}}{2}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ b.M_A=\dfrac{13}{0,2}=65\left(\dfrac{g}{mol}\right)\\ \rightarrow A:Kẽm\left(Zn=65\right)\\ c.n_{HCl}=2.n_A=0,4\left(mol\right)\\ V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\)
\(n_{H_2}=\dfrac{1,85925}{24,79}=0,075\left(mol\right)\)
\(m_{dd.HCl}=500.1,2=600\left(g\right)\)
\(2R+6HCl\rightarrow2RCl_3+3H_2\)
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.
\(m_{dd.X}=1,35+600-0,075.2=601,2\left(g\right)\)
\(n_{HCl.dư}=0,025.2=0,05\left(mol\right)\)
\(C\%_{AlCl_3}=\dfrac{0,05.133,5.100\%}{601,2}=1,11\%\)
\(C\%_{HCl.dư}=\dfrac{0,05.36,5.100\%}{601,2}=0,3\%\)
\(A+H_2O\rightarrow AOH+\dfrac{1}{2}H_2\)
Bảo toàn khối lượng => \(m_{H_2}=3,45+102,7-106=0,15\left(g\right)\)
=> \(n_A=2n_{H_2}=0,15\left(mol\right)\)
=>\(M_A=\dfrac{3,45}{0,15}=23\)
=> A là Na
\(C\%_{NaOH}=\dfrac{0,15.40}{106}.100=5,66\%\)
a) Đặt kim loại hóa trị II cần tìm là A.
\(n_{Ag}=\dfrac{75,6}{108}=0,7\left(mol\right)\\ A+2AgNO_3\rightarrow A\left(NO_3\right)_2+2Ag\\ n_A=\dfrac{n_{Ag}}{2}=\dfrac{0,7}{2}=0,35\left(mol\right)\\ M_A=\dfrac{m_A}{n_A}=\dfrac{19,6}{0,35}=56\left(\dfrac{g}{mol}\right)\)
Vậy kim loại A (II) cần tìm là sắt (Fe=56)
b)
\(n_{AgNO_3}=n_{Ag}=0,7\left(mol\right)\\ C_{MddAgNO_3}=\dfrac{0,7}{0,14}=5\left(M\right)\)
c)
\(V_{ddsau}=V_{ddAgNO_3}=0,14\left(l\right)\\ C_{MddFe\left(NO_3\right)_2}=\dfrac{0,35}{0,14}=2,5\left(M\right)\)