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a) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
nFe = nH2 = 0,3 (mol)
\(\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
b) nHCl = 2.nH2 = 0,6 (mol)
\(\Rightarrow V_{ddHCl}=\dfrac{0,6}{0,3}=2\left(l\right)\)
c) \(n_{FeCl_2}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow C_{M\left(FeCl_2\right)}=\dfrac{0,3}{2}=0,15M\)
a, \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(m_{HCl}=109,5.10\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
Mol: 0,1 0,3 0,1 0,15
Ta có: \(\dfrac{0,1}{2}=\dfrac{0,3}{6}\) ⇒ Al hết, HCl hết
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
b, \(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
c, mdd sau pứ = 2,7 + 109,5 - 0,15.2 = 111,9 (g)
\(C\%_{ddAlCl_3}=\dfrac{13,35.100\%}{111,9}=11,93\%\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\n_{HCl}=\dfrac{109,5\cdot10\%}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{2}=\dfrac{0,3}{6}\) \(\Rightarrow\) Al và HCl đều p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{AlCl_3}=0,1\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,1\cdot133,5=13,35\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Al}+m_{ddHCl}-m_{H_2}=111,9\left(g\right)\)
\(\Rightarrow C\%_{AlCl_3}=\dfrac{13,35}{111,9}\cdot100\%\approx11,93\%\)
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
_____0,4--->0,8------>0,4--->0,4
=> VH2 = 0,4.22,4 = 8,96(l)
c) mHCl = 0,8.36,5 = 29,2 (g)
=> \(m_{dd\left(HCl\right)}=\dfrac{29,2.100}{7,3}=400\left(g\right)\)
mdd (sau pư) = 22,4 + 400 - 0,4.2 = 421,6 (g)
=> \(C\%\left(FeCl_2\right)=\dfrac{127.0,4}{421,6}.100\%=12,05\%\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
____0,1______0,2_____0,1____0,1 (mol)
a, \(C_{M_{HCl}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}\left(M\right)\)
\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, \(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)
Theo PT: \(n_{NaOH}=2n_{FeCl_2}=0,2\left(mol\right)\)
\(\Rightarrow V_{NaOH}=\dfrac{0,2}{2}=0,1\left(l\right)\)
Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)
a. PTHH: Mg + H2SO4 ---> MgSO4 + H2↑
Theo PT: \(n_{H_2}=n_{Mg}=0,4\left(mol\right)\)
=> \(V_{H_2}=0,4.22,4=8,96\left(lít\right)\)
b. Theo PT: \(n_{H_2SO_4}=n_{Mg}=0,4\left(mol\right)\)
=> \(m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)
Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{39,2}{m_{dd_{H_2SO_4}}}.100\%=10\%\)
=> \(m_{dd_{H_2SO_4}}=392\left(g\right)\)
c. Ta có: \(m_{H_2}=0,4.2=0,8\left(g\right)\)
=> \(m_{dd_{MgSO_4}}=9,6+392-0,8=400,8\left(g\right)\)
Theo PT: \(n_{MgSO_4}=n_{Mg}=0,4\left(mol\right)\)
=> \(m_{MgSO_4}=0,4.120=48\left(g\right)\)
=> \(C_{\%_{MgSO_4}}=\dfrac{48}{400,8}.100\%=11,98\%\)
a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
x_____2x_____________x (mol)
\(Al_2\left(CO_3\right)_3+6HCl\rightarrow2AlCl_3+3H_2O+3CO_2\uparrow\)
y_____6y______________________3y (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}56x+234y=29\\x+3y=\dfrac{8,96}{22,4}=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=56\cdot0,1=5,6\left(g\right)\\m_{Al_2\left(CO_3\right)_3}=23,4\left(g\right)\end{matrix}\right.\)
b) Theo PTHH: \(n_{HCl}=2n_{Fe}+6n_{Al_2\left(CO_3\right)_3}=0,8\left(mol\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{0,8}{2}=0,4\left(l\right)=400\left(ml\right)\)
c) Kết tủa sau phản ứng không có Al(OH)3
Bảo toàn Sắt: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,05\left(mol\right)\) \(\Rightarrow m_{Fe_2O_3}=0,05\cdot160=8\left(g\right)\)
a:
\(n_{HCl}=0.2\cdot1=0.2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
b: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,1 0,2 0,1 0,1
\(m_{Fe}=0.1\cdot56=5.6\left(g\right)\)
c: \(m_{FeCl_2}=0.1\cdot\left(56+35.5\cdot2\right)=12.7\left(g\right)\)
d: \(V_{H_2}=0.1\cdot22.4=2.24\left(lít\right)\)
Bài 1 xem lại đề phần 2 nhé=)
3.
\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ PTHH:\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,05->0,1---->0,05---->0,05
a. \(V_{HCl}=\dfrac{0,1}{2}=0,05\left(l\right)\)
b. \(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
c. \(CM_{FeCl_2}=\dfrac{0,05}{0,05}=1M\)
`HaNa♬D`
1
\(1.Na_2O+H_2O\rightarrow2NaOH\\ 2.2KClO_3\xrightarrow[]{t^0}2KCl+3O_2\)
KCl ko ra đc AlCl3 nhé