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\(n_{Zn}=a\left(mol\right),n_{Al}=b\left(mol\right)\)
\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+H_2\)
\(n_{H_2}=a+1.5b=0.4\left(mol\right)\left(1\right)\)
\(m_{Muối}=m_{ZnCl_2}+m_{AlCl_3}=136a+133.5b=40.3\left(g\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.2\)
\(m_{hh}=0.1\cdot65+0.2\cdot27=11.9\left(g\right)\)
\(\%Zn=\dfrac{0.1\cdot65}{11.9}\cdot100\%=54.62\%\)
\(\%Al=100-54.62=45.38\%\)
3 kim loại + O → 3 oxit
...2,13 gam..........3,33 gam
=> lệch 3,33 – 2,13 = 1,2 gam = m (O trong oxit)
nO (trong ocid) = 1,2 / 16 =0.075 (mol)
Theo phản ứng : 2H + O = H2O =>số mol H = 0,075.2 = 0,15 mol
Thể tích HCl: 0,15 / 2 = 0,075 lít = 75 ml → Chọn C. 75 ml
mO = 3.33 - 2.13 = 1.2g -> nO = 0.075mol
Ta có: 2H+ + O-2 -> H2O
------- 0.15 <-0.075
-> V HCl = 0.15 :2 = 0.075 (l)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+3H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ a,n_{HCl}=2.n_{H_2}=2.0,4=0,8\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,8}{8}=0,1\left(l\right)\\ b,FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\downarrow\\ AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ n_{AgCl}=n_{AgNO_3}=3.n_{AlCl_3}+2.n_{FeCl_2}=3.a+2.b=3.0,2+2.0,1=0,8\left(mol\right)\\ \Rightarrow a=\dfrac{170.0,8}{250}.100=54,4\%\\ b=m_{\downarrow}=m_{AgCl}=0,8.143,5=114,8\left(g\right)\)
\(n_{HCl}=\dfrac{58,4.15\%}{36,5}=0,24\left(mol\right)\\ Fe+2HCl\rightarrow\left(t^o\right)FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{0,24}{2}=0,12\left(mol\right)\\ \Rightarrow V1=V_{H_2\left(đktc\right)}=0,12.22,4=2,688\left(l\right)\\ x=m_{Cu}=m_{hhA}-m_{Fe}=15,68-0,12.56=8,96\left(g\right)\\ b,n_{Cu}=\dfrac{8,96}{64}=0,14\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ Cu+Cl_2\rightarrow\left(t^o\right)CuCl_2\\ n_{Cl_2}=\dfrac{3}{2}.n_{Fe}+n_{Cu}=\dfrac{3}{2}.0,12+0,14=0,32\left(mol\right)\\ \Rightarrow V2=V_{Cl_2\left(đktc\right)}=0,32.22,4=7,168\left(l\right)\\ y=m_{muối}=m_{AlCl_3}+m_{CuCl_2}=0,12.133,5+0,14.135=34,92\left(g\right)\)
\(n_{MnO_2}=\frac{69,6}{87}=0,8\left(mol\right)\)
\(PTHH:MnO_2+4HCl_{\left(\text{đ}\right)}\underrightarrow{t^o}MnCl_2+2H_2O+Cl_2\)
(mol)_____0,8_____3,2________0,8_____1,6_____0,8__
\(V_{Cl_2}=0,8.22,4=17,92\left(l\right)\)
\(n_{NaOH}=0,25.2=0,5\left(mol\right)\)
\(PTHH:2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O\)
(mol)______0,5______0,25___0,25_____0,25______
Tỉ lệ: \(\frac{0,5}{2}< \frac{0,8}{1}\rightarrow Cl_2\) dư
\(C_{M\left(NaCl\right)}=C_{M\left(NaClO\right)}=\frac{0,25}{0,25}=1\left(M\right)\)
Câu 1:
a, \(n_{H2}=0,15\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1____0,3______0,1______0,15
\(m_{Al}=2,7\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\frac{2,7.100}{12,9}=20,93\%\\\%m_{Al2O3}=100\%-20,93\%=76,07\%\end{matrix}\right.\)
b, \(m_{Al2O3}=12,9-2,7=10,2\left(g\right)\)
\(\Rightarrow n_{Al2O3}=0,1\left(mol\right)\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
0,1_____0,6________0,2___________
\(\Sigma n_{HCl}=0,9\left(mol\right)\Rightarrow m_{HCl}=32,85\left(g\right)\)
\(\Rightarrow m_{dd\left(HCl\right)}=32,85:10\%=298,63\left(mol\right)\)
\(V_{HCl}=\frac{328,5}{1,1}=298,63\left(ml\right)=0,29863\left(l\right)\)
\(n_{AlCl3}=0,3\left(mol\right)\)
\(\Rightarrow CM_{AlCl3}=1\left(M\right)\)
Câu 2:
Coi muối gồm kim loại và nhóm Cl
\(\Rightarrow m_{Cl}=32,53-7,5=25,03\left(g\right)\)
\(n_{HCl}=n_{Cl}=\frac{25,03}{35,5}=0,7\left(mol\right)\)
\(\Rightarrow m=m_{dd\left(HCl\right)}=0,7.36,5:10\%=255,5\left(g\right)\)