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a) 2Al + 6HCl -> 2AlCl3 + 3H2
Al2O3 + 6HCl -> 2AlCl3 + 3H2O
nH2 = 0,15mol => nAl=0,1mol => mAl=2,7g; mAl2O3 = 10,2g => nAl2O3 = 0,1mol
=>%mAl=20,93% =>%mAl2O3 = 79,07%
b) nHCl = 0,1.3+0,1.6=0,9 mol=>mHCl(dd)=100g
mddY=12,9+100-0,15.2=112,6g
mAlCl3=22,5g=>C%=19,98%
Fe+2HCl->FeCl2+H2
x---2x-----------x
Mg+2HCl->MgCl2+H2
y------2y-----------y
Ta có :
\(\left\{{}\begin{matrix}56x+24y=24\\x+y=\dfrac{13,44}{22,4}\end{matrix}\right.\)
=>x=0,3 mol, y=0,3 mol
=>%m Fe=\(\dfrac{0,3.56}{24}.100\)=70%
=>%m Mg=100-70=30%
=>VHCl=\(\dfrac{0,3.2+0,3.2}{2}\)=0,6l=600ml
b)
XCl2+2AgNO3->2AgCl+X(NO3)2
0,6--------------------1,2mol
=>m AgCl=1,2.143,5=172,2g
1) Ptpư:
2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
Fe + 2HCl \(\rightarrow\) FeCl2 + H2
Cu + HCl \(\rightarrow\) không phản ứng
=> 0,6 gam chất rắn còn lại chính là Cu:
Gọi x, y lần lượt là số mol Al, Fe
Ta có:
3x + 2y = 2.0,06 = 0,12
27x + 56 y = 2,25 – 0,6 = 1,65
=> x = 0,03 (mol) ; y = 0,015 (mol)
=> \(\%Cu=\frac{0,6}{2,25}.100\%=26,67\%\); \(\%Fe=\frac{56.0,015}{2,25}.100\%=37,33\%\); %Al = 36%
2) \(n_{SO_2}=\frac{1,344}{22,4}=0,06mol\); m (dd KOH) = 13,95.1,147 = 16 (gam)
=> mKOH = 0,28.16 = 4,48 (gam)=> nKOH = 0,08 (mol)=> \(1<\)\(\frac{n_{KOH}}{n_{SO_2}}<2\)
=> tạo ra hỗn hợp 2 muối: KHSO3: 0,04 (mol) và K2SO3: 0,02 (mol)
Khối lượng dung dịch sau pu = 16 + 0,06.64 = 19,84 gam
=> \(C\%\left(KHSO_3\right)=\frac{0,04.120}{19,84}.100\%\)\(=24,19\%\)
\(C\%\left(K_2SO_3\right)=\frac{0,02.158}{19,84}.100\%\)\(=15,93\%\)
\(n_{H_2}=\dfrac{7,84}{22,4}=0,35mol\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Zn}=y\end{matrix}\right.\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
x x ( mol )
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}56x+65y=21,4\\x+y=0,35\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,2\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,15.56=8,4g\)
\(\Rightarrow m_{Zn}=0,2.65=13g\)
\(\%m_{Fe}=\dfrac{8,4}{21,4}.100=39,25\%\)
\(\%m_{Zn}=100\%-39,25\%=60,75\%\)
\(m_{FeCl_2}=0,15.127=19,05g\)
\(m_{ZnCl_2}=0,2.136=27,2g\)
Bài 1:
\(n_{HCl}=2.0,16=0,32\left(mol\right);n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
PTHH: Fe + 2HCl → FeCl2 + H2
\(m_{H_2}=0,16.2=0,32\left(g\right)\)
\(m_{HCl}=0,32.36,5=11,68\left(g\right)\)
Theo ĐLBTKL ta có: \(m_{MgCl_2+FeCl_2}=1,4+11,68-0,32=12,76\left(g\right)\)
Bài 12:
Theo ĐLBTKL, ta có:
\(m_{hhkl}+m_{O_2}=m_{hh.oxit}\\ \Leftrightarrow11,9+m_{O_2}=18,3\\ \Leftrightarrow m_{O_2}=18,3-11,9=6,4\left(g\right)\\ n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
\(a,Fe+2HCl\rightarrow FeCl_2+H_2\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\Rightarrow n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Fe}=\dfrac{0,1.56}{13,6}.100\%\approx41,176\%\\ \Rightarrow\%m_{CuO}\approx58,824\%\\ b,n_{CuO}=\dfrac{13,6-0,1.56}{80}=0,1\left(mol\right)\\ n_{HCl\left(p.ứ\right)}=2.\left(n_{Fe}+n_{CuO}\right)=2.\left(0,1+0,1\right)=0,4\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)
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 )
2.
a)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{H2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
Gọi a là số mol Fe b là số mol Zn\(\left\{{}\begin{matrix}56a+65b=29,8\\a+b=0,5\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\)
\(\%m_{Fe}=\frac{0,3.56}{29,8}.100\%=56,38\%\)
\(\%m_{Zn}=100\%=56,38\%=43,62\%\)
b)
\(n_{HCl}=0,5.2=1\left(mol\right)\)
\(CM_{HCl}=\frac{1}{0,6}=\frac{5}{2}M\)
1.
a,\(2Fe+3Cl_2\underrightarrow{^{to}}2FeCl_3\)
\(n_{Fe}=\frac{1,68}{56}=0,03\left(mol\right)\)
\(n_{Cl2}=\frac{0,84}{22,4}0,0375\left(mol\right)\)
Lập tỉ lệ : \(\frac{n_{Fe}}{2}=0,015< \frac{n_{CL2}}{3}=0,0125\)
Vậy Cl2 hết Fe dư
\(n_{Fe_{du}}=n_{Fe}-\frac{2}{3}n_{Cl2}\)
\(=0,03-\frac{2}{3}.0,0375=0,005\left(mol\right)\)
\(\rightarrow m_{Fe_{du}}=0,005.56=0,28\left(g\right)\)
\(n_{FeCl3}=\frac{2}{3}n_{Cl2}=0,025\left(mol\right)\)
\(\rightarrow m_{FeCl3}=0,025.162,5=4,065\left(g\right)\)
b, \(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,005___0,01__________
\(\rightarrow n_{HCl}=0,01\left(mol\right)\)
\(V_{HCl}=\frac{0,01}{0,5}=0,02\left(l\right)\)