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M tb hh = 27,5 => hỗn hợp có NH3 , vì hh X có 2 chất HC pư với NaOH tạo khí có 2C => khí còn lại là CH3NH2 => X có CH3COONH4 và HCOOCH3NH3 . Pư :
CH3COONH4 + NaOH ---------> CH3COONa + NH3 + H2O
a a a
HCOOCH3NH3 + NaOH -------> HCOONa + CH3NH2 + H2O
b b b
ta có n hh = a + b = 0,2 mol
m hh = Mtb.n = 5,5 = 17a + 31b
từ hệ => a = 0,05 , b = 0,15 mol => m muối khan = 0,05.82 + 0,15 . 68 = 14,3g => B
\(a.2Al+6HCl->2AlCl_3+3H_2\\ Mg+2HCl->MgCl_2+H_2\\ b.n_{Al}=a,n_{Mg}=b\\ 27a+24b=7,5\left(I\right)\\ 1,5a+b=\dfrac{7,84}{22,4}=0,35\left(II\right)\\ a=0,1;b=0,2\\ \%m_{Al}=\dfrac{27\cdot0,1}{7,5}\cdot100\%=36\%\\ \%m_{Mg}=64\%\\ c.m_{HCl}=36,5\left(0,1\cdot3+0,2\cdot2\right)=18,25g\\ d.m_{ddsau}=7,5+\dfrac{18,25}{14,6:100}-0,35\cdot2=131,8g\\ C\%\left(AlCl_3\right)=\dfrac{133,5\cdot0,1}{131,8}\cdot100\%=10,1\%\\ C\%\left(MgCl_2\right)=\dfrac{95\cdot0,2}{131,8}\cdot100\%=14,4\%\)
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??
2KMnO4--->K2MnO4+MnO2+O2 n KMnO4=15,8/158=0,1(mol) n O2=1/2n KMnO4=0,05(mol) V O2=0,05.22,4=1,12(l)
Câu 2: a) SO3+H2O--->H2SO4 b) m H2SO4=20.10/100=2(g) n H2SO4=2/98=0,02(mol) n SO3=n H2SO4=0,02(mol) m =m SO3=0,02.80=1,6(g)
Câu 3 : a) Fe+2HCl-->FeCl2+H2 x--------------------------x(mol) Mg+2HCl------->MgCl2+H2 y------------------------------y(mol) n H2=4,48/22,4=0,2(mol) Theo bài ra ta có hpt \(\left\{{}\begin{matrix}56x+24y=8\\x+y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\) n Fe : n Al= 1 : 1
Câu 4: a) Hiện tượng : có chất rắn màu nâu đỏ sau pư PT: FeCl3+3KOH--->3KCl+Fe(OH)3 b) m KOH=\(\frac{200.8,4}{100}=16,8\left(g\right)\) n KOH=16,8/56=0,3(mol) n Fe(OH)3=1/3n KOH=0,1(mol) m Fe(OH)3=0,1.107=10,7(g) c) n FeCl3=1/3n KOH=0,1(mol) m FeCl3=0,1.162,5=16,25(g) m dd FeCl3=16,25.100/6,5=250(g) m dd sau pư=m FeCl3+m dd KOH- m Fe(OH)3 =250+200-10,7=439,3(g) n KCl=n KOH=0,3(mol) m KCl=74,5.0,3=22,35(g) C% KCl=22,35/439,3.100%=5,09% d) 2Fe(OH)3--->Fe2O3+3H2O n Fe2O3=1/2n Fe(OH)3=0,05(mol) a=m Fe2O3=0,05.160=8(g)
Câu 5: a) Al2O3+6HCl---->2Alcl3+3H2O x----------6x(mol) MgO+2HCl----->MgCl2+H2O y-----------2y(mol) m HCl=90.7,3/100=6,57(g) n HCl=6,57/36,5=0,18(mol) Theo bài ta có hpt \(\left\{{}\begin{matrix}102x+40y=3,24\\6x+2y=0,18\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,03\end{matrix}\right.\) %m Al2O3=\(\frac{0,02.102}{3,24}.100\%=62,96\%\) %m MgO=100-62,96=37,04% b)m dd sau pư=m KL+m dd HCl=3,24+90=93,24(g) m AlCl3=0,04.133,5=5,34(g) C% Alcl3=5,34/92,24.100%=5,79% m MgCl2=0,03.95=2,85(g) C% MgCl2=2,85/92,24.100%=2,8%
Câu 6: a) n H2=1,456/22,4=0,056(mol) 2Al+3H2SO4---.Al2(SO4)3+3H2 x-------------------------------1,5x Fe+H2SO4--->FeSO4+H2 y--------------------------------y(mol) Theo bài ra ta có hpt \(\left\{{}\begin{matrix}27x+56y=1,93\\1,5x+y=0,065\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,03\\y=0,02\end{matrix}\right.\) %m Al=0,03.27/1,93.100%=41,97% %m Fe=100-41,97=53,08% c) 2Al+3Cu(NO3)2---->3Cu+2Al(NO3) 0,03------------------------0,045(mol) Fe+Cu(NO3)2---->Fe(NO3)2+Cu 0,02-------------------------------0,02(mol) m Cu=(0,045+0,02).64=4,16(g)
Câu 7: a) Zn+2HCl---.Zncl2+H2 b) n H2=3,36/22,4=0,15(mol) n Zn=n H2=0,15(mol) m Zn=0,15.65=9,75(g) %m Zn=9,75/10,05.100%=97% %m Cu=3%
Câu 9:
Gọi oxit KL Cần tìm là MO
MO+H2SO4--->MSO4+H2O
n H2SO4=19,6/98=0,2(mol)
n MO=n H2SO4=0,2(mol)
M MO=16/0,2=80
M+18=80-->M=64(Cu)
Vậy M là Cu
\(a,n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Ba + 2H2O ---> Ba(OH)2 + H2
0,3<-------------0,3<---------0,3
=> mBa = 0,3.137 = 41,1 (g)
=> mK2O = 59,9 - 41,1 = 18,8 (g)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Ba}=\dfrac{41,1}{59,9}.100\%=68,61\%\\\%m_{K_2O}=100\%-68,61\%=31,39\%\end{matrix}\right.\)
\(b,n_{K_2O}=\dfrac{18,8}{94}=0,2\left(mol\right)\)
PTHH: K2O + H2O ---> 2KOH
0,2----------------->0,4
Các chất tan trong dd sau phản ứng: KOH, Ba(OH)2
\(\rightarrow\left\{{}\begin{matrix}m_{KOH}=0,4.56=22,4\left(g\right)\\m_{Ba\left(OH\right)_2}=0,3.171=51,3\left(g\right)\end{matrix}\right.\)
Gọi \(\left\{{}\begin{matrix}n_{NaOH}=a\left(mol\right)\\n_{KOH}=b\left(mol\right)\end{matrix}\right.\)
\(n_{Mg\left(OH\right)_2}=\dfrac{14,5}{58}=0,25\left(mol\right)\)
PTHH:
2NaOH + MgSO4 ---> Mg(OH)2 + Na2SO4
a -----------------------------> 0,5a
2KOH + MgSO4 ---> Mg(OH)2 + K2SO4
b -------------------------------> 0,5b
Hệ pt \(\left\{{}\begin{matrix}40a+56b=24,8\\0,5a+0,5b=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,3\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,2.40=8\left(g\right)\\m_{KOH}=0,3.56=16,8\left(g\right)\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}\%m_{NaOH}=\dfrac{8}{24,8}=32,26\%\\\%m_{KOH}=100\%-32,26\%=67,74\%\end{matrix}\right.\)
2NaOH+MgSO4->Mg(OH)2+Na2SO4
x-----------------------------1\2x
2KOH+MgSO4->K2SO4+Mg(OH)2
y--------------------------------------1\2y
=> ta có :
\(\left\{{}\begin{matrix}40x+56y=24,8\\0,5x+0,5y=0,25\end{matrix}\right.=>\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)
%mNaOH=\(\dfrac{0,2.40}{24,8}100\)=32,25%
=>%m KOH=67,75%