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\(n_{NaOH}=1.0,4=0,4(mol);n_{FeCl_3}=1.0,1=0,1(mol)\\ a,PTHH:3NaOH+FeCl_3\to Fe(OH)_3\downarrow+3NaCl\\ \text {Vì }\dfrac{n_{NaOH}}{3}>\dfrac{n_{FeCl_3}}{1} \text {nên }NaOH\text { dư}\\ \Rightarrow n_{Fe(OH)_3}=0,1(mol)\\ \Rightarrow m_{Fe(OH)_3}=107.0,1=10,7(g)\\ b,n_{NaCl}=3n_{FeCl_3}=0,3(mol)\\ \Rightarrow C_{M_{NaCl}}=\dfrac{0,3}{0,4+0,1}=0,6M\)
Bài 19 :
\(a) n_{Al} = \dfrac{10,8}{27} = 0,4(mol)\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,6(mol)\\ V_{H_2} = 0,6.22,4 = 13,44(lít)\\ b) \text{Chất tan : }Al_2(SO_4)_3\\ n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,2(mol)\\ m_{Al_2(SO_4)_3} = 0,2.342 = 68,4(gam)\)
Bài 18 :
\(a) n_{HCl} = \dfrac{250.7,3\%}{36,5 } = 0,5(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,25(mol) \Rightarrow V_{H_2} = 0,25.22,4 = 5,6(lít)\\ b) \text{Chất tan : } ZnCl_2\\ n_{ZnCl_2} = n_{H_2} = 0,25(mol)\\ m_{ZnCl_2} = 0,25.136 = 34(gam)\)
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, \(BaCl_2+H_2SO_4\rightarrow2HCl+BaSO_{4\downarrow}\)
b, \(m_{BaCl_2}=200.2,08\%=4,16\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{4,16}{208}=0,02\left(mol\right)\)
\(m_{H_2SO_4}=300.9,8\%=29,4\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,02}{1}< \dfrac{0,3}{1}\), ta được H2SO4 dư.
Theo PT: \(\left\{{}\begin{matrix}n_{BaSO_4}=n_{H_2SO_4\left(pư\right)}=n_{BaCl_2}=0,02\left(mol\right)\\n_{HCl}=2n_{BaCl_2}=0,04\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,3-0,02=0,28\left(mol\right)\)
Ta có: m dd sau pư = m dd BaCl2 + m dd H2SO4 - mBaSO4 = 200 + 300 - 0,02.233 = 495,34 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,04.36,5}{495,34}.100\%\approx0,295\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,28.98}{495,34}.100\%\approx5,54\%\end{matrix}\right.\)
a, Gọi \(m_{NaCl\left(thêm\right)}=a\left(g\right)\)
\(m_{NaCl\left(bđ\right)}=5\%.100=5\left(g\right)\\ \Rightarrow C\%_{NaCl}=\dfrac{5+a}{100+a}.100\%=5,5\%\\ \Leftrightarrow a=0,53\left(g\right)\)
b, \(m_{NaCl}=58,5.5,5\%=3,2175\left(g\right)\\ n_{NaCl}=\dfrac{3,2175}{58,5}=0,055\left(mol\right)\)
PTHH: NaCl + AgNO3 ---> AgCl↓ + NaNO3
0,055-->0,055------>0,055---->0,055
\(m_{AgCl}=0,055.143,5=7,8925\left(g\right)\\ m_{ddY}=58,5+200-7,8925=250,6075\left(g\right)\\ \Rightarrow C\%_{NaNO_3}=\dfrac{0,055.85}{250,6075}.100\%=1,87\%\)
\(n_{KOH}=\dfrac{400.7\%}{56}=0,5\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.19,6\%}{98}=0,2\left(mol\right)\)
PTHH: 2KOH + H2SO4 --> K2SO4 + 2H2O
Xét tỉ lệ: \(\dfrac{0,5}{2}>\dfrac{0,2}{1}\) => KOH dư, H2SO4 hết
PTHH: 2KOH + H2SO4 --> K2SO4 + 2H2O
0,4<----0,2-------->0,2
=> \(\left\{{}\begin{matrix}m_{KOH\left(dư\right)}=\left(0,5-0,4\right).56=5,6\left(g\right)\\m_{K_2SO_4}=0,2.174=34,8\left(g\right)\end{matrix}\right.\)
mdd sau pư = 400 + 100 = 500 (g)
=> \(\left\{{}\begin{matrix}C\%_{KOH.dư}=\dfrac{5,6}{500}.100\%=1,12\%\\C\%_{K_2SO_4}=\dfrac{34,8}{500}.100\%=6,96\%\end{matrix}\right.\)
\(n_{KOH}=\dfrac{400.7}{100}:56=0,5\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.19,6}{100}:98=0,2\left(mol\right)\)
\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
0,4 0,2 0,2
Lập tỉ lệ:
\(\dfrac{0,5}{2}>\dfrac{0,2}{1}\) => KOH dư.
\(m_{dd}=400+100=500\left(g\right)\)
\(n_{KOH.dư}=0,5-0,4=0,1\left(mol\right)\)
\(C\%_{K_2SO_4}=\dfrac{0,2.174.100}{500}=6,96\%\)
\(C\%_{KOH}=\dfrac{0,1.56.100}{500}=1,12\%\)
\(\begin{cases} m_{H_2SO_4}=\dfrac{100.19,6\%}{100\%}=19,6(g)\\ m_{BaCl_2}=\dfrac{300.20,8\%}{100\%}=62,4(g) \end{cases} \Rightarrow \begin{cases} n_{H_2SO_4}=\dfrac{19,6}{98}=0,2(mol)\\ n_{BaCl_2}=\dfrac{62,4}{208}=0,3(mol) \end{cases}\\ a,PTHH:BaCl_2+H_2SO_4\to BaSO_4\downarrow +2HCl\)
Vì \(\dfrac{n_{H_2SO_4}}{1}<\dfrac{n_{BaCl_2}}{1}\) nên \(BaCl_2\) dư
\(\Rightarrow n_{BaSO_4}=0,2(mol)\\ \Rightarrow m_{BaSO_4}=0,2.233=46,6(g)\)
\(b,n_{HCl}=n_{BaSO_4}=0,2(mol)\\ \Rightarrow m_{CT_{HCl}}=0,2.36,5=7,3(g)\\ m_{dd_{HCl}}=100+300-46,6=353,4(g)\\ \Rightarrow C\%_{HCl}=\dfrac{7,3}{353,4}.100\%\approx 2,07\%\)