Cho 2.8 gam kim loại R (II) phản ứng vừa đủ với dung dịch AgNO3 5% thu được 9 gam muối của R. a/ Tìm R b/ Tính khối lượng dung dịch AgNO3 phản ứng c/ Tính nồng độ % chất tan trong dung dịch sau phản ứng
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![](https://rs.olm.vn/images/avt/0.png?1311)
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a, Gọi \(m_{NaCl\left(thêm\right)}=a\left(g\right)\)
\(m_{NaCl\left(bđ\right)}=5\%.100=5\left(g\right)\\ \Rightarrow C\%_{NaCl}=\dfrac{5+a}{100+a}.100\%=5,5\%\\ \Leftrightarrow a=0,53\left(g\right)\)
b, \(m_{NaCl}=58,5.5,5\%=3,2175\left(g\right)\\ n_{NaCl}=\dfrac{3,2175}{58,5}=0,055\left(mol\right)\)
PTHH: NaCl + AgNO3 ---> AgCl↓ + NaNO3
0,055-->0,055------>0,055---->0,055
\(m_{AgCl}=0,055.143,5=7,8925\left(g\right)\\ m_{ddY}=58,5+200-7,8925=250,6075\left(g\right)\\ \Rightarrow C\%_{NaNO_3}=\dfrac{0,055.85}{250,6075}.100\%=1,87\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) nH2=0,15(mol)
PTHH: R + 2 HCl -> RCl2 + H2
0,15______0,3____0,15___0,15(mol)
M(R)=mR/nR=3,6/0,15= 24(g/mol)
=> R(II) cần tìm là Magie (Mg=24)
b) PTHH: Mg + 2 HCl -> MgCl2 + H2
mHCl=0,3.36,5=10,95(g)
=>C%ddHCl= (10,95/150).100= 7,3%
c) mH2= 0,15.2=0,3(g)
mddMgCl2= mMg + mddHCl - mH2= 3,6+ 150 - 0,3= 153,3(g)
mMgCl2=0,15.95=14,25(g)
=> \(C\%ddMgCl2=\dfrac{14,25}{153,3}.100\approx9,295\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
\(PTHH:R+2HCl\rightarrow RCl_2+H_2\uparrow\\ Mol:0,2\leftarrow0,4\rightarrow0,2\rightarrow0,2\)
=> MR = \(\dfrac{13}{0,2}=65\left(\dfrac{g}{mol}\right)\)
=> R là Zn
=> \(\left\{{}\begin{matrix}V_{H_2}=0,2.22,4=4,48\left(l\right)\\m_{ZnCl_2}=0,2.136=27,2\left(g\right)\end{matrix}\right.\)
nHCL = 14,6 : 36,5 = 0,4 (MOL)
pthh : 2R + 2xHCl ---> 2RClx + xH2
0,4x<--0,4 (mol)
MR = 13:0,4x = 32,5x(g/mol)
xét
x = 1 (KTM )
x= 2 (TM )
x = 3 (KTM )
x =4( KTM )
x= 5 (ktm )
x=6 (ktm)
x=7 (ktm )
=> R là zn
![](https://rs.olm.vn/images/avt/0.png?1311)
mik sửa lại cái dưới bị lỗi latex
\(a.n_{HCl}=0,05.2=0,1\left(mol\right);n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\ 2R+6HCl\rightarrow2RCl_3+3H_2\\ \Rightarrow\dfrac{0,1}{6}>\dfrac{0,03}{3}\Rightarrow HCl.dư,R.pư.hết\\ n_R=0,03.2:3=0,02\left(mol\right)\\ M_R=\dfrac{0,54}{0,02}=27\left(g/mol\right)\\ \Rightarrow R=27\left(Al,nhôm\right)\\ b.C_{M_{AlCl_3}}=\dfrac{0,3.2:3}{0,05}=0,4M\\ C_{M_{HCl\left(dư\right)}}=\dfrac{0,1-\left(0,3.6:3\right)}{0,05}=0,8M\)
\(a.n_{HCl}=0,05.2=0,1\left(mol\right)\\ n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\ 2R+6HCl\rightarrow2RCl_3+3H_2\\ \Rightarrow\dfrac{0,1}{6}>\dfrac{0,03}{3}\Rightarrow HCl.dư,R.pư.hết\\ n_R=0,03.2:3=0,02\left(mol\right)\\ M_R=\dfrac{0,54}{0,02}=27\left(g/mol\right)\\ \Rightarrow R=27\left(Al,nhôm\right)\\ b.n_{AlCl_3}=n_{Al}=0,02mol\\ C_{M_{AlCl_3}}=\dfrac{0,02}{0,05}=0,4M\\ C_M_{HCl\left(dư\right)}=\dfrac{0,1-\left(0,03.2\right)}{0,05}=0,8M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=8,3\\ 1,5a+b=\dfrac{5,6}{22,4}=0,25\\ a=b=0,1\\ m_{Al}=27\cdot0,1=2,7g\\ m_{Fe}=8,3-2,7=5,6g\\ a=\dfrac{3a+2b}{500}\cdot36,5=3,65\%\\ m_{ddsau}=508,3-0,25\cdot2=507,8g\\ C\%_{AlCl_3}=\dfrac{133,5a}{507,8}=2,63\%\\ C\%_{FeCl_2}=\dfrac{127b}{507,8}=2,50\%\)
\(a,PTHH:R+2AgNO_3\to R(NO_3)_2+2Ag\\ \Rightarrow n_{R}=n_{R(NO_3)_2}\\ \Rightarrow \dfrac{2,8}{M_R}=\dfrac{9}{M_R+124}\\ \Rightarrow M_R=56(g/mol)\)
Vậy R là sắt (Fe)
\(b,n_{R}=\dfrac{2,8}{56}=0,05(mol)\\ \Rightarrow n_{AgNO_3}=0,1(mol)\\ \Rightarrow m_{dd_{AgNO_3}}=\dfrac{0,1.170}{5\%}=340(g)\\ c,n_{Fe(NO_3)_2}=n_{Fe}=0,05(mol);n_{Ag}=0,1(mol)\\ \Rightarrow C\%_{Fe(NO_3)_2}=\dfrac{0,05.180}{2,8+340-0,1.108}.100\%=2,71\%\)
Dạ em cảm ơn ạ