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a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(Cu+2H_2SO_{4\left(đ\right)}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O\)
b, Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
\(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)
\(n_{SO_2}=\dfrac{2,9748}{24,79}=0,12\left(mol\right)\)
\(n_{Cu}=n_{SO_2}=0,12\left(mol\right)\)
\(\Rightarrow m=m_{Zn}+m_{Cu}=0,1.65+0,12.64=14,18\left(g\right)\)
Có: \(n_{H_2SO_{4\left(đ\right)}}=2n_{SO_2}=0,24\left(mol\right)\Rightarrow x=m_{ddH_2SO_4\left(đ\right)}=\dfrac{0,24.98}{98\%}=24\left(g\right)\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right);n_{SO_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
PTHH:
Fe + 2HCl ---> FeCl2 + H2
0,2------------------------>0,2
\(2Fe+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
0,2---------------------------------------->0,3
\(Cu+2H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}CuSO_4+SO_2+2H_2O\)
0,15<--------------------------------0,15
=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,2.56}{0,2.56+0,15.64}.100\%=53,85\%\\\%m_{Cu}=100\%-53,85\%=46,15\%\end{matrix}\right.\)
a.Mg + H2SO4 -> MgSO4 + H2
b.\(nH2=\dfrac{4.704}{22.4}=0.21mol\) = nMg
mMg = 0.21\(\times24=5.04g\)
\(\%mMg=\dfrac{5.04\times100}{25}=20.16\%\)
\(\%mAg=100-20.16=79.84\%\)
c.MgSO4 + 2KOH -> K2SO4 + Mg(OH)2
0.21 0.42
H2SO4 + 2KOH -> K2SO4 + H2O
0.04 0.08
\(nH2SO4=\dfrac{9.8\times250}{100\times98}=0.25mol\)
Mà nH2SO4 phản ứng = nH2 = 0.21 mol
\(\Rightarrow nH2SO4dư=0.25-0.21=0.04mol\)
=> nKOH = 0.42 + 0.08 = 0.5mol
\(\Rightarrow CM_{KOH}=\dfrac{0.5}{0.625}=0.8M\)
Quy đổi hh ban đầu thành \(\left\{{}\begin{matrix}Fe:a\left(mol\right)\\S:b\left(mol\right)\end{matrix}\right.\)
=> 56a + 32b = 3,2 (1)
PTHH: 2Fe + 6H2SO4 --> Fe2(SO4)3 + 3SO2 + 6H2O
a------->3a--------->0,5a
S + 2H2SO4 --> 3SO2 + 2H2O
b---->2b
=> 3a + 2b = 0,18 (2)
(1)(2) => a = 0,04 (mol); b = 0,03 (mol)
=> \(n_{Fe_2\left(SO_4\right)_3}=0,02\left(mol\right)\)
=> mFe2(SO4)3 = 0,02.400 = 8 (g)
Quy đổi hh ban đầu thành \(\left\{{}\begin{matrix}Fe:a\left(mol\right)\\S:b\left(mol\right)\end{matrix}\right.\)
=> 56a + 32b = 3,2 (1)
PTHH: 2Fe + 6H2SO4 --> Fe2(SO4)3 + 3SO2 + 6H2O
a------->3a--------->0,5a
S + 2H2SO4 --> 3SO2 + 2H2O
b---->2b
=> 3a + 2b = 0,18 (2)
(1)(2) => a = 0,04 (mol); b = 0,03 (mol)
=> \(n_{Fe_2\left(SO_4\right)_3}=0,02\left(mol\right)\)
=> mFe2(SO4)3 = 0,02.400 = 8 (g)
\(\left(a\right)Zn+2HCl\rightarrow ZnCl_2+H_2\\ DungdịchX:ZnCl_2, A:H_2,B:Ag\\ n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ n_{HCl}=2n_{H_2}=0,5\left(mol\right)\\ \Rightarrow x=m_{ddHCl}=\dfrac{0,5.36,5}{3,65}=500\left(g\right)\\ n_{Zn}=n_{H_2}=0,2\left(mol\right)\\ \Rightarrow y=m_{Ag}=27,05-0,2.65=14,05\left(g\right)\\ \left(b\right):m_{ddsaupu}=0,2.65+500-0,2.2=512,6\left(g\right)\\ TheoPT:n_{ZnCl_2}=n_{H_2}=0,2\left(mol\right)\\ C\%_{ZnCl_2}=\dfrac{0,2.136}{512,5}.100=5,3\%\)
Qui đổi ½ hh B gồm Al (x mol), Fe (y mol), O (z mol)
=> mB = 2 (mAl + mFe + mO) = 102,78g
Gọi công thức của oxit sắt là FeaOb
=> Fe2O3
a, \(Cu+2H_2SO_{4\left(đ\right)}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O\)
b, \(n_{SO_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
Theo PT: \(n_{Cu}=n_{CuSO_4}=n_{SO_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,15.64=9,6\left(g\right)=m\)
Theo PT: \(n_{H_2SO_4}=2n_{SO_2}=0,3\left(mol\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{0,3.98}{200}.100\%=14,7\%=x\)
Ta có: m dd sau pư = 9,6 + 200 - 0,15.64 = 200 (g)
\(\Rightarrow C\%_{CuSO_4}=\dfrac{0,15.160}{200}.100\%=12\%\)