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Câu 1:
Gọi : nMg=a(mol); nMgO=b(mol) (a,b>0)
a) PTHH: Mg + 2 HCl -> MgCl2 + H2
a________2a_______a______a(mol)
MgO +2 HCl -> MgCl2 + H2O
b_____2b_______b___b(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}24a+40b=8,8\\22,4a=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
=> mMg=0,2.24=4,8(g)
=>%mMg= (4,8/8,8).100=54,545%
=> %mMgO= 45,455%
b) m(muối)=mMg2+ + mCl- = 0,3. 24 + 0,6.35,5=28,5(g)
c) V=VddHCl=(2a+2b)/2=0,3(l)=300(ml)
Câu 2:
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{HCl}=0,4\cdot2=0,8\left(mol\right)\end{matrix}\right.\)
PTHH: \(Ca+2HCl\rightarrow CaCl_2+H_2\uparrow\)
0,2____0,4_____0,2____0,2 (mol)
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
0,2____0,4______0,2____0,2 (mol)
Ta có: \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,2\cdot40}{0,2\cdot40+0,2\cdot56}\cdot100\%\approx41,67\%\\\%m_{CaO}=58,33\%\\m_{CaCl_2}=\left(0,2+0,2\right)\cdot111=44,4\left(g\right)\end{matrix}\right.\)
1.GS có 100g dd $HCl$
=>m$HCl$=100.20%=20g
=>n$HCl$=20/36,5=40/73 mol
=>n$H2$=20/73 mol
Gọi n$Fe$(X)=a mol n$Mg$(X)=b mol
=>n$HCl$=2a+2b=40/73
mdd sau pứ=56a+24b+100-40/73=56a+24b+99,452gam
m$MgCl2$=95b gam
C% dd $MgCl2$=11,79%=>95b=11,79%(56a+24b+99,452)
=>92,17b-6,6024a=11,725
=>a=0,13695 mol và b=0,137 mol
=>C%dd $FeCl2$=127.0,13695/mdd.100%=15,753%
2.Bảo toàn klg=>mhh khí bđ=m$C2H2$+m$H2$
=0,045.26+0,1.2=1,37 gam
mC=mA-mbình tăng=1,37-0,41=0,96 gam
HH khí C gồm $H2$ dư và $C2H6$ không bị hấp thụ bởi dd $Br2$ gọi số mol lần lượt là a và b mol
Mhh khí=8.2=16 g/mol
mhh khí=0,96=2a+30b
nhh khí=0,06=a+b
=>a=b=0,03 mol
Vậy n$H2$=n$C2H6$=0,03 mol
Câu 6:
a+b) Ta có: \(\left\{{}\begin{matrix}n_{khí}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\\overline{M}_{khí}=11,5\cdot2=23\end{matrix}\right.\)
Theo sơ đồ đường chéo: \(n_{H_2}=n_{CO_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\)
\(\Rightarrow\%V_{H_2}=\%V_{CO_2}=50\%\)
PTHH: \(M+2HCl\rightarrow MCl_2+H_2\uparrow\)
0,1____0,2_____0,1___0,1 (mol)
\(MCO_3+2HCl\rightarrow MCl_2+CO_2\uparrow+H_2O\)
0,1____0,2_____0,1____0,1_____0,1 (mol)
Ta có: \(0,1M+0,1\left(M+60\right)=10,8\) \(\Leftrightarrow M=24\)
Vậy kim loại cần tìm là Magie
c) Theo các PTHH: \(n_{HCl\left(p/ứ\right)}=0,4\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,4\cdot105\%=0,42\left(mol\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{0,42}{2}=0,21\left(l\right)=210\left(ml\right)\)
Câu 5: a) \(M_Y=11,5.2=23\)
CaCO3 + 2HNO3 → Ca(NO3)2 + CO2 + H2O
Hỗn hợp Y gồm CO2 và 1 khí khác
Vì trong Y M CO2 > MY
=> M khí còn lại phải < M Y
=> Khí còn lại là H2 ( Do các sản phẩm khử của HNO3 đều có M > MY)
Ca + 2HNO3 -------> Ca(NO3)2 + H2
Gọi x, y lần lượt là số mol của CaCO3, Ca
=> n CO2 = x ; nH2 = y
Theo đề bài ta có hệ phương trình sau :
\(\left\{{}\begin{matrix}x+y=0,1\\44x+2y=23.0,1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)
=> \(m_{CaCO_3}=0,05.100=5\left(g\right)\)
\(m_{Ca}=0,05.40=2\left(g\right)\)
=> \(\%m_{CaCO_3}=\dfrac{5}{5+2}.100=71,43\%\)
=> \(\%m_{Ca}=100-71,43=28,57\%\)
b) Dung dịch Y gồm Ca(NO3)2
\(\Sigma n_{Ca\left(NO_3\right)_2}=0,05+0,05=0,1\left(mol\right)\)
\(m_{ddHNO_3}=\dfrac{\left(0,05.2+0,05.2\right).63}{6,3\%}=200\left(g\right)\)
\(m_{ddsaupu}=7+200-2,3=204,7\left(g\right)\)
=> \(C\%_{Ca\left(NO_3\right)_2}=\dfrac{0,1.164}{204,7}.100=8,01\%\)
ta có 200cm3=0,2 lítn hcl=2*0,2=0,4 molgọi số mol của caco3 là a,na2co3 là bcaco3 + 2hcl -> cacl2 + co2 + h2oa(mol)---2a(mol)--a-------a--------ana2co3 + 2hcl -> 2nacl + co2 + h2ob(mol)---2b(mol)---2b-------b------bta có100a+106b=20,62a+2b=0,4=> a=b=0,1 mol=> m caco3=10g; m na2co3=10,6 g
PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\) (1)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\) (2)
a) Ta có: \(\Sigma n_{HCl}=0,2\cdot2=0,4\left(mol\right)\)
Gọi số mol của Na2CO3 là \(a\) \(\Rightarrow n_{HCl\left(1\right)}=2a\left(mol\right)\)
Gọi số mol của CaCO3 là \(b\) \(\Rightarrow n_{HCl\left(2\right)}=2b\left(mol\right)\)
Ta lập được hệ phương trình:
\(\left\{{}\begin{matrix}2a+2b=0,4\\106a+100b=20,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,1\cdot100=10\left(g\right)\\m_{Na_2CO_3}=10,6\left(g\right)\end{matrix}\right.\)
b) Theo PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=2n_{Na_2CO_3}=0,2mol\\n_{CaCl_2}=n_{CaCO_3}=0,1mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\\m_{CaCl_2}=0,1\cdot111=11,1\left(g\right)\end{matrix}\right.\)
Mặt khác: \(\left\{{}\begin{matrix}m_{CO_2}=0,2\cdot44=8,8\left(g\right)\\m_{ddHCl}=200\cdot1,2=240\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd\left(saup/ư\right)}=m_{hh}+m_{ddHCl}-m_{CO_2}=251,8\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{11,7}{251,8}\cdot100\%\approx4,65\%\\C\%_{CaCl_2}=\dfrac{11,1}{251,8}\cdot100\%\approx4,41\%\end{matrix}\right.\)
\(a.n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ a............3a.......a.........1,5a\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b.........2b........b.........b\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}27a+56b=5,5\\1,5a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.0,1}{5,5}.100\approx49,091\%\\\%m_{Fe}=\dfrac{0,05.56}{5,5}.100\approx50,909\%\end{matrix}\right.\\ b.C_{MddHCl}=\dfrac{3a+2b}{0,5}=\dfrac{3.0,1+2.0,05}{0,5}=0,8\left(M\right)\)
\(a,\) Đặt \(n_{Al}=x(mol);n_{Fe}=y(mol)\)
\(\Rightarrow 27x+56y=11(1)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Fe+H_2SO_4\to FeSO_4+H_2\\ Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3\downarrow+3Na_2SO_4\\ FeSO_4+2NaOH\to Fe(OH)_2\downarrow+Na_2SO_4\\ \Rightarrow n_{Al(OH)_3}=x;n_{Fe(OH)_2}=y\\ \Rightarrow 78x+90y=24,6(2)\\ (1)(2)\Rightarrow \begin{cases} x=0,2(mol)\\ y=0,1(mol) \end{cases} \Rightarrow \begin{cases} m_{Al}=0,2.27=5,4(g)\\ m_{Fe}=11-5,4=5,6(g) \end{cases}\)
\(b,\Sigma n_{H_2SO_4}=1,5x+y=0,4(mol)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{0,4}{0,2}=2(l)\\ c,\Sigma n_{NaOH}=3x+2y=0,8(mol)\\ \Rightarrow m_{dd_{NaOH}}=\dfrac{0,8.40}{10\%}=320(g)\\ d,2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\\ Fe(OH)_2\xrightarrow{t^o}FeO+H_2O\\ \Rightarrow n_{Al_2O_3}=0,1(mol);n_{FeO}=0,1(mol)\\ \Rightarrow m_{\text{chất rắn}}=0,1.102+0,1.72=17,4(g)\)
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\)
\(n_{HCl}=0,2\cdot4=0,8mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(x\) \(\rightarrow\) \(3x\) \(x\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(y\) \(\rightarrow\) \(2y\) \(y\)
\(\Rightarrow\left\{{}\begin{matrix}27x+65y=11,9\\3x+2y=0,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
a)\(\%m_{Al}=\dfrac{0,2\cdot27}{11,9}\cdot100\%=45,38\%\)
\(\%m_{Zn}=100\%-45,38\%=54,62\%\)
b)\(\Sigma n_{H_2}=\dfrac{3}{2}x+y=\dfrac{3}{2}\cdot0,2+0,1=0,4mol\)
\(V_{H_2}=0,4\cdot22.4=8,96l\)
a.
\(m_{Ag}=m_{k.tan}=8,7\left(g\right)\\ m_{Zn,Mg}=20-8,7=11,3\left(g\right)\\ \left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \rightarrow\left\{{}\begin{matrix}65a+24b=11,3\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Ag}=\dfrac{8,7}{20}.100=43,5\%\\\%m_{Mg}=\dfrac{24.0,2}{20}.100=24\%\\\%m_{Zn}=\dfrac{0,1.65}{20}.100=32,5\%\end{matrix}\right.\)
b.
\(n_{H_2SO_4\left(tổng\right)}=a+b=0,3\left(mol\right)\\ V_{ddH_2SO_4\left(tổng\right)}=\dfrac{0,3}{0,5}=0,6\left(lít\right)=600\left(ml\right)\)
