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Câu 1:
\(Zn\rightarrow Zn^{+2}+2e\)
\(2H^++2e\rightarrow H_2^o\)
Ta có:
\(m_{Zn}=\frac{9,45}{65}=0,15\left(mol\right)\)
Bảo toàn e:
\(2n_{Zn}=2n_{H2}\Rightarrow n_{H2}=n_{Zn}=0,15\left(mol\right)\)
Câu 2:
\(Al\rightarrow Al^{+3}+3e\)
\(2H^++2e\rightarrow H_2^o\)
Ta có:
\(n_{H2}=\frac{10,80}{22,4}=0,45\left(mol\right)\)
Bảo toàn e:
\(3n_{Al}=2n_{H2}\)
\(\Rightarrow n_{Al}=\frac{2}{3}n_{H2}=0,3\left(mol\right)\)
Câu 3:
\(Cu\rightarrow Cu^++2e\)
\(S^{+6}+2e\rightarrow S^{+4}\)
Ta có:
\(n_{Cu}=\frac{19,2}{64}=0,3\left(mol\right)\)
Bảo toàn e:
\(2n_{Cu}=2n_{SO2}\Rightarrow n_{SO2}=n_{Cu}=0,3\left(mol\right)\)
Câu 4:
\(Zn\rightarrow Zn^{+2}+2e\)
\(Al\rightarrow Al^{+3}+3e\)
\(O_2+4e\rightarrow2O^{-2}\)
Gọi \(\left\{{}\begin{matrix}n_{Zn}:x\left(mol\right)\\n_{Al}:y\left(mol\right)\end{matrix}\right.\)
Bảo toàn e:
\(2n_{Zn}+3n_{Al}=4n_{O2}\)
\(2x+3y=1\Leftrightarrow x=y=0,2\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=13\left(g\right)\\m_{Al}=5,4\left(g\right)\end{matrix}\right.\)
Câu 5:
\(Mg\rightarrow Mg^{+2}+2e\)
\(Fe\rightarrow Fe^{+3}+3e\)
\(S^{+6}+2e\rightarrow S^{+4}\)
Bảo toàn e:
\(2n_{Mg}+3n_{Fe}=2n_{SO2}\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
a_____a_________________ a
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b_____ b ___________________b
Giải hệ PT:
\(\left\{{}\begin{matrix}56a+65b=12,1\\a+b=\frac{4,48}{22,4}=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\frac{0,1.56}{12,1}.100\%=46,28\%\\\%m_{Zn}=100\%-46,28\%=53,72\%\end{matrix}\right.\)
\(\Rightarrow V_{H2SO4\left(can.dung\right)}=\frac{0,1+0,1}{0,2}=0,1\left(l\right)\)
\(\Rightarrow\left\{{}\begin{matrix}CM_{FeSO4}=\frac{0,1}{0,1}=1M\\CM_{ZnSO4}=\frac{0,1}{0,1}=1M\end{matrix}\right.\)
Zn:0
H2CO3:
H:+1
C:+4
O:-2
CuO:
Cu:+2
O:-2
H2:0
*-2 nhé : HPO4:
H:+1
P:+5
O:-2
Ca2+:+2
NaNO3:
Na:+1
N:+5
O:-2
Ca(NO3)2:
Ca:+2
N:+5
O:-2
MgCl2:
Mg:+2
Cl:-1
PO4(-3):
P:+5
O:-2
K+:+1
NO3-:
N:+5
O:-2
SO3:
S:+6
O:-2
N2O5:
N:+5
O:-2
NH4NO3:
N:-3
H:+1
N:+5
O:-2
Fe2O3:
Fe:+3
O:-2
KClO4:
K:+1
Cl:+7
O:-2
Fe2(SO4)3:
Fe:+2
S:+6
O:-2
P2O3:
P:+3
O:-2
a, \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Gọi a là số mol của Fe2O3 , b là số mol của CuO
\(\Rightarrow\frac{160a.100}{160a+80b}=80\)
\(\Rightarrow16000a=12800a+6400b\Rightarrow a-2b=0\left(1\right)\)
\(n_{HCl}=0,14\left(mol\right)\Rightarrow6a+2b=0,14\left(2\right)\)
\(\left(1\right)+\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=0,02\\b=0,01\end{matrix}\right.\)
\(\Rightarrow m=m_{Fe2O3}+m_{CuO}=4\left(g\right)\)
b,
\(m_{FeCl3}=162,5.2a=6,5\left(g\right)\)
\(m_{CuCl2}=135.b=1,35\left(g\right)\)
\(V_{dd_{HCl}}=140\left(ml\right)\Rightarrow m_{dd}=140.1,2=168\left(g\right)\)
\(\Rightarrow m_{dd_{spu}}=168+4=172\left(g\right)\)
\(C\%_{FeCl3}=\frac{6,5.100}{172}=3,78\%\)
\(C\%_{CuCl2}=\frac{1,35.100}{172}=0,78\%\)
a,
\(Fe+H_2O\rightarrow FeO+H_2\)
\(3Fe+4H_2O\rightarrow Fe_3O_4+4H_2\)
\(20Fe+6KNO_3\rightarrow5Fe_2O_3+3K_2O+3N_2\)
b,
\(2KNO_3\underrightarrow{^{to}}2KNO_2+O_2\)
\(4P+5O_2\underrightarrow{^{to}}2P_2O_5\)
\(P_2O_5+3H_2O\underrightarrow{^{to}}2H_3PO_4\)
\(4Fe+3H_3PO_4\rightarrow Fe_3\left(PO_4\right)_2+FeHPO_4+4H_2\)
\(3Fe+4H_2O\rightarrow Fe_3O_4+4H_2\)
\(3Fe_3O_4+8H_3PO_4\rightarrow Fe_3\left(PO_4\right)_2+FePO_4+12H_2O\)
Vì số mol FeO bằng số mol Fe2O3 nên ta quy hỗn hợp về Fe3O4
\(n_{Fe3O4}=\frac{2,32}{232}=0,1\left(mol\right)\)
PTHH :
Fe3O4 + 8HCl ---> FeCl2 +2FeCl3 + 4H2O
.0,1...........0,8...........0,1..........0,2...........0,4
\(Vdd_{HCl}=0,8\cdot1=0,8\left(l\right)\)
Vậy V = 0,8
Đkc hay đktc vậy nhìn nó hơi rối :v