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\(n_{H2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
a 0,4 0,2 1a
\(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
b 0,3 0,15 1b
a) Gọi a là số mol của Mg
b là số mol của Fe
\(m_{Mg}+m_{Fe}=13,2\left(g\right)\)
⇒ \(n_{Mg}.M_{Mg}+n_{Fe}.M_{Fe}=13,2g\)
⇒ 24a + 56b = 13,2g (1)
Theo phương trình : 1a + 1b = 0,35(2)
Từ(1),(2), ta có hệ phương trình :
24a + 56b = 13,2g
1a + 1b = 0,35
⇒ \(\left\{{}\begin{matrix}a=0,2\\b=0,15\end{matrix}\right.\)
\(m_{Mg}=0,2.24=4,8\left(g\right)\)
\(m_{Fe}=0,15.56=8,4\left(g\right)\)
0/0Mg = \(\dfrac{4,8.100}{13,2}=36,36\)0/0
0/0Fe = \(\dfrac{8,4.100}{13,2}=63,64\)0/0
b) \(n_{HCl\left(tổng\right)}=0,4+0,3=0,7\left(mol\right)\)
200ml = 0,2l
\(C_{M_{ddHCl}}=\dfrac{0,7}{0,2}=3,5\left(M\right)\)
c) \(m_{muối.clorua}=\left(0,2.95\right)+\left(0,15.127\right)=38,05\left(g\right)\)
Chúc bạn học tốt
\(\left\{{}\begin{matrix}Fe\\Al\end{matrix}\right.+HCl->\left\{{}\begin{matrix}FeCl2\\AlCl3\end{matrix}\right.+H2\)
Ta có số mol Fe là x , Al là y (mol)
\(\left\{{}\begin{matrix}56x+27y=11\\127x+133,5y=39,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%mFe=\dfrac{0,1.56}{11}=50,9\%\\\%mAl=\dfrac{0,2.27}{11}=49,09\%\end{matrix}\right.\)
Bảo toàn e :
\(2.nH2=2.nFe+3.nAl\Rightarrow nH2=0,4\left(mol\right)\)
\(V=0,4.22,4=8,96\left(l\right)\)
\(nFe=nFeCl2=0,1\left(mol\right)\)
\(nAl=nAlCl3=0,2\left(mol\right)\)
\(\Rightarrow nHCl\left(pứ\right)=2.0,1+3.0,2=0,8\left(mol\right)\)
\(Cm=\dfrac{n}{V}=\dfrac{0,8}{0,2}=4\left(M\right)\)
\(a,Fe+2HCl\rightarrow FeCl_2+H_2\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\Rightarrow n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Fe}=\dfrac{0,1.56}{13,6}.100\%\approx41,176\%\\ \Rightarrow\%m_{CuO}\approx58,824\%\\ b,n_{CuO}=\dfrac{13,6-0,1.56}{80}=0,1\left(mol\right)\\ n_{HCl\left(p.ứ\right)}=2.\left(n_{Fe}+n_{CuO}\right)=2.\left(0,1+0,1\right)=0,4\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)
a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Ta có: \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
Theo PT: \(n_{Mg}=n_{H_2}=0,03\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,03.24=0,72\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,72}{1,74}.100\%\approx41,38\%\\\%m_{AlCl_3}\approx58,62\%\end{matrix}\right.\)
b, Theo PT: \(n_{HCl}=2n_{H_2}=0,06\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,06.36,5=2,19\left(g\right)\)
\(\Rightarrow C\%_{ddHCl}=\dfrac{2,19}{500}.100\%=0,438\%\)
Bạn tham khảo nhé!
a, nH2 = 0,03 ( mol )
=> nMg = nH2 = 0,03 ( mol )
=> mMg = 0,72 g
=> %Mg \(\approx\) 41,38 % .
=> % Al \(\approx\) 58,62 % .
b, Có : nH2 = 0,03 mol
=> nHCl = nHCltừ Al2O3 + nHCltừ Mg = 0,06 + 0,06 = 0,12 ( mol )
=> mHCl = 4,38 ( g )
Lại có : mdd = mhh + mddHCl = 501,74 ( g )
=> \(C\%=\dfrac{m_{HCl}}{m_{dd}}.100\%\approx0,87\%\)
( chắc đoạn trên là Al2O3 :vvvv )
a, Ta có: 24nMg + 56nFe = 9,2 (g) (1)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
BT e, có: 2nMg + 2nFe = 2nH2 = 0,5 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,15\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,15.24}{9,2}.100\%\approx39,13\%\\\%m_{Fe}\approx60,87\%\end{matrix}\right.\)
b, BTNT H, có: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)
\(n_{HCl}=\dfrac{58,4.15\%}{36,5}=0,24\left(mol\right)\\ Fe+2HCl\rightarrow\left(t^o\right)FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{0,24}{2}=0,12\left(mol\right)\\ \Rightarrow V1=V_{H_2\left(đktc\right)}=0,12.22,4=2,688\left(l\right)\\ x=m_{Cu}=m_{hhA}-m_{Fe}=15,68-0,12.56=8,96\left(g\right)\\ b,n_{Cu}=\dfrac{8,96}{64}=0,14\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ Cu+Cl_2\rightarrow\left(t^o\right)CuCl_2\\ n_{Cl_2}=\dfrac{3}{2}.n_{Fe}+n_{Cu}=\dfrac{3}{2}.0,12+0,14=0,32\left(mol\right)\\ \Rightarrow V2=V_{Cl_2\left(đktc\right)}=0,32.22,4=7,168\left(l\right)\\ y=m_{muối}=m_{AlCl_3}+m_{CuCl_2}=0,12.133,5+0,14.135=34,92\left(g\right)\)
\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.1......0.2....................0.1\)
\(m_{Cu}=m_{hh}-m_{Fe}=12-0.1\cdot56=6.4\left(g\right)\)
\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)