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PTHH :
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,1 0,2 0,1 0,1
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,09 0,18 0,09 0,09
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(a,\%m_{Mg}=\dfrac{0,1.24}{6}=40\%\)
\(\%m_{MgO}=100\%-40\%=60\%\)
\(n_{MgO}=\dfrac{6-2,4}{40}=0,09\left(mol\right)\)
\(b,m_{HCl}=\left(0,2+0,18\right).36,5=13,87\left(g\right)\)
\(m_{ddHCl}=\dfrac{13,87.100}{20}=69,35\left(g\right)\)
\(V_{ddHCl}=\dfrac{m}{D}=\dfrac{69,35}{1,1}\approx63\left(ml\right)\) ( cái này mình nghĩ đề phải là D bạn nhé tại vì khối lượng riêng của HCl là 1,18g/ml )
\(c,m_{MgCl_2}=\left(0,09+0,1\right).95=18,05\left(g\right)\)
\(m_{ddMgCl_2}=6+69,35-0,2=75,15\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{18,05}{75,15}.100\%\approx24,02\%\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
a) Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)=n_{Mg}\)
\(\Rightarrow\%m_{Mg}=\dfrac{0,5\cdot24}{16}=75\%\) \(\Rightarrow\%m_{MgO}=25\%\)
b) Ta có: \(\left\{{}\begin{matrix}n_{Mg}=0,5\left(mol\right)\\n_{MgO}=\dfrac{16\cdot25\%}{40}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl}=2n_{Mg}+2n_{MgO}=1,2\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{1,2\cdot36,5}{20\%}=219\left(g\right)\)
c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{H_2}=0,5\left(mol\right)\\n_{MgCl_2}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2}=0,5\cdot2=1\left(g\right)\\m_{MgCl_2}=0,6\cdot95=57\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{hhA}+m_{ddHCl}-m_{H_2}=234\left(g\right)\) \(\Rightarrow C\%_{MgCl_2}=\dfrac{57}{234}\cdot100\%\approx24,36\%\)
Cho mình hỏi ở cái PTHH ấy! sao ta không tính số mol ở dưới??
dạ em cảm ơn anh/thầy nhưng mà cái tổng HCl ra m bấm máy sai rồi ạ vs cảm ơn anh/thầy giúp em giải bài nha
a) nH2 = \(\dfrac{2,24}{22,4}=0,1\) mol
Pt: Mg + 2HCl --> MgCl2 + ....H2
0,1 mol<-0,2 mol<-0,1 mol<-0,1 mol
mMg = 0,1 . 24 = 2,4 (g)
mCu = mhh - mMg = 3,04 - 2,4 = 0,64 (g)
% mMg = \(\dfrac{2,4}{3,04}.100\%=78,95\%\)
% mCu = \(\dfrac{0,64}{3,04}.100\%=21,05\%\)
b) mHCl = 0,2 . 36,5 = 7,3 (g)
=> mdd HCl = \(\dfrac{7,3\times100}{20}=36,5\left(g\right)\)
c) mMgCl2 = 0,1 . 95 = 9,5 (g)
mH2 = 0,1 . 2 = 0,2 (g)
mdd sau pứ = mMg + mdd HCl - mH2 = 2,4 + 36,5 - 0,2 = 38,7 (g)
C% dd MgCl2 = \(\dfrac{9,5}{38,7}.100\%=24,55\%\)
a, \(n_{H_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,8\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,8}{0,5}=1,6\left(l\right)\)
b, Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\) ⇒ 27x + 24y = 3,78 (1)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Mg}=\dfrac{3}{2}x+y=0,4\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=\\y=\end{matrix}\right.\)
Đến đây thì ra số mol âm, bạn xem lại đề nhé.
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{109,5.20\%}{36,5}=0,6\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,2}{1}< \dfrac{0,6}{2}\Rightarrow HCldư\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ n_{HCl\left(dư\right)}=0,6-0,2.2=0,2\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,m_{ZnCl_2}=136.0,2=27,2\left(g\right)\\ c,m_{ddsau}=13+109,5-0,2.2=122,1\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{27,2}{122,1}.100\approx22,277\%\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,2.36,5}{122,1}.100\approx5,979\%\)
Zn + 2HCl -> ZnCl2 + H2
a, nZn = 13/65= 0,2(mol)
mHCl= 109,5.20%/100%=21.9(g)
nHCl=21,9/36,5=0,6(mol)
Theo PT nHCl = 2nZn= 2.0,2= 0,4(mol)<0,6(mol)
=> HCl pư dư, Zn pư hết
Theo PT: nH2= nZn =0,2(mol)
VH2=0,2.22,4=4,48(l)
b, Theo PT: nZnCl2=nZn=0,2(mol)
mZnCl2= 0,2.136=27,2(g)
c, mdd sau pư= 13+109,5-0,2.2=122,1(g)
C%dd ZnCl2=27,2.100%/122,1=22,28%
nHCl dư= 0,6-0,4=0,2(mol)
mHcl
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
\(n_{Fe}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
\(\Rightarrow m_{Cu}=20-5,6=14,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{20}.100\%=28\%\\\%m_{Cu}=72\%\end{matrix}\right.\)
Câu 2 :
\(n_{Cu}=a\left(mol\right),n_{Al}=b\left(mol\right)\)
\(m=64a+27b=11.8\left(g\right)\left(1\right)\)
\(BTKL:m_{O_2}=18.2-11.8=6.4\left(g\right)\)
\(n_{O_2}=\dfrac{6.4}{32}=0.2\left(mol\right)\)
\(2Cu+O_2\underrightarrow{^{^{t^0}}}2CuO\)
\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)
\(n_{O_2}=0.5a+0.75b=0.2\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.2\)
\(\%Cu=\dfrac{0.1\cdot64}{11.8}\cdot100\%=54.23\%\)
PTHH: Mg + 2HCl \(\rightarrow\) MgCl2 + H2\(\uparrow\) (1)
Theo pt: 1......... 2 ........... 1 ........ 1 ... (mol)
Theo đề: 0,1 ... 0,2 ........ 0,1 ..... 0,1 .. (mol)
PTHH: MgO + 2HCl \(\rightarrow\) MgCl2 + H2O (2)
Theo pt: . 1......... 2 ............ 1 ......... 1 ... (mol)
Theo đề: 0,09 ... 0,18 ...... 0,09 .... 0,09 . (mol)
1/ \(n_{H_2}=\dfrac{V_{đktc}}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(m_{Mg}=n.M=0,1.24=2,4\left(g\right)\)
Ta có: mMg + mMgO = 6 (gt)
\(\Rightarrow\) mMgO = 6 - mMg = 6 - 2,4 = 3,6 (g)
\(\%m_{Mg}=\dfrac{m_{Mg}}{m_{hh}}.100\%=\dfrac{2,4}{6}.100\%=40\%\)
\(\%m_{MgO}=\dfrac{m_{MgO}}{m_{hh}}.100\%=\dfrac{3,6}{6}.100\%=60\%\)
2/ \(n_{MgO}=\dfrac{m}{M}=\dfrac{3,6}{40}=0,09\left(mol\right)\)
\(n_{HCl}=n_{HCl\left(1\right)}+n_{HCl\left(2\right)}=0,2+0,18=0,38\left(mol\right)\)
\(m_{HCl}=n.M=0,38.36,5=13,87\left(g\right)\)
\(m_{ddHCl}=\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{13,87.100\%}{20\%}=69,35\left(g\right)\)
\(V_{ddHCl}=\dfrac{m_{ddHCl}}{D}=\dfrac{69,35}{1,12}\approx61,92\left(ml\right)\approx0,062\left(l\right)\)
3/ \(m_{MgCl_2}=m_{MgCl_{2\left(1\right)}}+m_{MgCl_{2\left(2\right)}}=95\left(0,1+0,09\right)=18,05\left(g\right)\)
\(m_{H_2}=n.M=0,1.2=0,2\left(g\right)\)
\(m_{ddspứ}=m_{Mg}+m_{MgO}+m_{ddHCl}-m_{H_2}=6+69,35-0,2=75,15\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{m_{MgCl_2}}{m_{ddspứ}}.100\%=\dfrac{18,05}{75,15}.100\%\approx24,02\%\)