Giải chi tiết giúp mình
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(2Fe+6H_2SO_4\text{(đặc)}\xrightarrow[]{t^\circ}Fe_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)
b) \(Fe+H_2SO_4\text{(loãng)}\xrightarrow[]{}FeSO_4+H_2\uparrow\)
c) \(\left\{{}\begin{matrix}2Na+2H_2O\xrightarrow[]{}2NaOH+H_2\uparrow\\NaOH+NH_4Cl\xrightarrow[]{}NaCl+NH_3\uparrow+H_2O\end{matrix}\right.\)
d) \(CuSO_4+BaCl_2\rightarrow BaSO_4\downarrow+CuCl_2\)
e) \(NaClO+HCl\text{(loãng)}\xrightarrow[]{}NaCl+HClO\)
g) \(NaClO+2HCl\text{(đặc)}\rightarrow NaCl+Cl_2\uparrow+H_2O\)
h) \(\left\{{}\begin{matrix}AlCl_3+3LiOH\xrightarrow[]{}3LiCl+Al\left(OH\right)_3\downarrow\\Al\left(OH\right)_3+LiOH\rightarrow LiAlO_2+2H_2O\end{matrix}\right.\)
a) \(C_2H_4+Br_2\xrightarrow[]{}C_2H_4Br_2\)
b) \(n_{C_2H_4Br_2}=n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(m_{C_2H_4Br_2}=0,5.188=94\left(g\right)\)
c) \(n_{Br_2}=n_{C_2H_4}=0,5\left(mol\right)\)
500ml=0,5l
\(C_{M_{Br_2}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
a) \(Mg+H_2SO_4\xrightarrow[]{}MgSO_4+H_2\\ MgO+H_2SO_4\xrightarrow[]{}MgSO_4+H_2O\)
b) \(n_{H_2SO_4}=\dfrac{200.24,5}{100.98}=0,5\left(mol\right)\)
Gọi x, y là số mol Mg, MgO
⇒ \(n_{H_2SO_4}=x+y=0,5\left(mol\right)\)
Ta có hpt: \(\left\{{}\begin{matrix}24x+40y=16,8\\x+y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)
\(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
Theo BTKL: \(m_{hh}+m_{ddH_2SO_4}=m_{ddsau}+m_{H_2}\)
⇒ \(m_{ddsau}=16,8+200-0,2.2=216,4\left(g\right)\)
\(m_{MgSO_4}=120.\left(0,2+0,3\right)=60\left(g\right)\)
\(C\%_{MgSO_4}=\dfrac{60}{216,4}.100=27,73\%\)
Câu 1:
1)
`(1) Fe + 2HCl -> FeCl_2 + H_2`
`(2) FeCl_2 + 2KOH -> Fe(OH)_2 + 2KCl`
`(3) Fe(OH)_2 + H_2SO_4 -> FeSO_4 + 2H_2O`
`(4) FeSO_4 + BaCl_2 -> FeCl_2 + BaSO_4`
2) Sửa `Fe(OH)_3 -> Fe(OH)_2`
`(1) Fe + 2HCl -> FeCl_2 + H_2`
`(2) FeCl_2 + 2KOH -> Fe(OH)_2 + 2KCl`
`(3) 4Fe(OH)_2 + O_2 -> (t^o) 2Fe_2O_3 + 4H_2O`
`(4) Fe_2O_3 + 3CO -> (t^o) 2Fe + 3CO_2`
3)
`(1) 2Al + 6HCl -> 2AlCl_3 + 3H_2`
`(2) AlCl_3 + 3KOH -> Al(OH)_3 + 3KCl`
`(3) 2Al(OH)_3 -> (t^o) Al_2O_3 + 3H_2O`
`(4) 2Al_2O_3 -> (đpnc, criolit) 4Al + 3O_2`
4)
`(1) 2Fe + 3Cl_2 -> (t^o) 2FeCl_3`
`(2) FeCl_3 + 3KOH -> Fe(OH)_3 + 3KCl`
`(3) 2Fe(OH)_3 -> (t^o) Fe_2O_3 + 3H_2O`
`(4) Fe_2O_3 + 3CO -> (t^o) 2Fe + 3CO_2`
Câu 2:
a) \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: Fe + H2SO4 ---> FeSO4 + H2
0,6<----0,6<-------0,6<----0,6
b) \(m_{muối}=m_{FeSO_4}=0,6.152=91,2\left(g\right)\)
c) \(C_{M\left(H_2SO_4\right)}=\dfrac{0,6}{0,3}=2M\)
Câu 3:
a) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
0,2<--0,3<------------0,1<--------0,3
b) \(m_{muối}=m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)
c) \(C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,3}=1M\)
a, Mg+H2SO4 -> MgSO4+H2
0,2 0,2 0,2 0,2
b, VH2=0,2.22,4=4,48L
c, VH2SO4= \(\dfrac{0.2}{2}\)=0,1L
d, MgSO4 + 2NaOH -> Mg(OH)2+Na2SO4
0,2 0,4
->mctNaOH= 16g
-> 5%=\(\dfrac{16}{mddNaOH}\).100% -> mddNaOH= 320g
a) \(Mg+H_2SO_4\xrightarrow[]{}MgSO_4+H_2\)
(mol) 0,2.......0,2............0,2...............0,2
b) \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
c) \(V_{H_2SO_4}=\dfrac{0,2}{2}=0,1\left(l\right)\)
d) \(2NaOH+MgSO_4\xrightarrow[]{}Na_2SO_4+Mg\left(OH\right)_2\)
(mol) 0,4.............0,2..................................0,2
\(m_{ddNaOH}=\dfrac{0,4.40.100}{5}=320\left(g\right)\)
\(m_{Mg\left(OH\right)_2}=0,2.58=11,6\left(g\right)\)
\(Fe\left\{{}\begin{matrix}FeCl_2\rightarrow FeCl_3\rightarrow Fe\left(OH\right)_2\\FeCl_3\rightarrow FeCl_2\rightarrow Fe\left(OH\right)_3\end{matrix}\right.\)
vậy được không bạn?
bạn ơi sơ đồ tư duy mà bạn cộng thêm đây chỉ là một hợp chất vô cơ thôi bạn ='((
1. \(n_{K_2CO_3}=\dfrac{13,8}{138}=0,1\left(mol\right)\)
PTHH: \(K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\)
0,1------>0,2------->0,2----->0,1
2. \(m_{ddHCl}=\dfrac{0,2.36,5}{20\%}=36,5\left(g\right)\)
3. \(V_{khí}=V_{CO_2}=0,1.22,4=2,24\left(l\right)\)
4. \(m_{ddspư}=36,5+13,8-0,1.44=45,9\left(g\right)\)
=> \(C\%_{KCl}=\dfrac{0,2.74,5}{45,9}.100\%=32,46\%\)
Câu 6 :
\(n_{K_2CO_3}=\dfrac{13.8}{138}=0,1\left(mol\right)\)
PTHH : K2CO3 + 2HCl -> 2KCl + CO2 \(\uparrow\) + H2O
0,1 0,2 0,2 0,1
\(m_{dd_{HCl}}=0,2.36,5=7,3\left(g\right)\)
\(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)
\(C\%_{dd_{KCl}}=\dfrac{0,2.74,5}{13,8+7,3-\left(0,1.44\right)}.100=89,22\%\)