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\(a)n_{Mg} = a(mol) ; n_{Fe} = b(mol) \Rightarrow 24a + 56b =2 0(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = a + b =\dfrac{11,2}{22,4} = 0,5(2)\\ (1)(2) \Rightarrow a = b = 0,25\\ \%m_{Mg} = \dfrac{0,25.24}{20}.100\% = 30\%\\ \%m_{Fe} = 100\%-30\% = 70\%\\ b) \\Mg^0 \to Mg^{2+} + 2e;Fe^0 \to Fe^{3+} + 3e\\ S^{+6} \to S^{+4} + 2e\\ 2n_{Mg} + 3n_{Fe} = 2n_{SO_2}\)
\(n_{SO_2} = \dfrac{0,25.2 + 0,25.3}{2} = 0,625(mol)\\ V_{SO_2} = 0,625.22,4 = 14(lít)\)
a, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: Fe + H2SO4 ---> FeSO4 + H2
0,2<---------------------------0,2
\(\rightarrow\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Cu}=16-11,2=4,8\left(g\right)\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{32}{16}.0,2=0,4\left(mol\right)\\n_{Cu}=\dfrac{4,8}{64}.\dfrac{32}{16}=0,15\left(mol\right)\end{matrix}\right.\)
PTHH:
Cu + 2H2SO4 (đặc, nóng) ---> CuSO4 + SO2 + 2H2O
0,15--------------------------------------------->0,15
2Fe + 6H2SO4 (đặc, nóng) ---> Fe2(SO4)3 + 3SO2 + 6H2O
0,4------------------------------------------------------>0,6
=> VSO2 = (0,6 + 0,15).22,4 = 16,8 (l)
c, \(n_{NaOH}=0,375.2=0,75\left(mol\right)\)
\(T=\dfrac{0,75}{0,6+0,15}=1\) => tạo duy nhất muối axit (NaHSO3)
PTHH: NaOH + SO2 ---> NaHSO3
0,75----------------->0,75
=> mmuối = 0,75.104 = 78 (g)
a/nH2= 0,1(mol)
Fe + H2SO4 -> FeSO4 + H2
0,1_________________0,1(mol)
=> mFe=0,1.56=5,6(g)
=> %mFe= (5,6/12).100\(\approx\) 46,667%
=> %mCu \(\approx\) 100% - 46,667% \(\approx\) 53,333%
b) mCu= 12-5,6=6,4(g) -> nCu= 0,1(mol)
Cu + 2 H2SO4(đ) -to-> CuSO4 + SO2 + 2 H2O
0,1___0,2__________________0,1(mol)
V=V(SO2,đktc)=0,1.22,4=2,24(l)
mH2SO4(p.ứ)=0,2.98=19,6(g)
=> mH2SO4(bđ)= 19,6 x 100/90 \(\approx21,778\left(g\right)\)
=> mddH2SO4 \(\approx\) (21,778 x 100)/98\(\approx22,222\left(g\right)\)
\(n_{Cu}=a\left(mol\right),n_{Fe}=b\left(mol\right)\)
\(m_X=64a+56b=16.2\left(g\right)\left(1\right)\)
\(n_{SO_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
Bảo toàn e :
\(2a+3b=0.4\cdot2=0.8\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.0475,b=0.235\)
\(\%Cu=\dfrac{0.0475\cdot64}{16.2}\cdot100\%=18.76\%\)
\(\%Fe=81.24\%\)
\(b.\)
\(\dfrac{a}{b}=\dfrac{0.0475}{0.235}=\dfrac{19}{94}\)
\(\Rightarrow n_{Cu}=19x\left(mol\right),n_{Fe}=94x\left(mol\right)\)
\(m_X=19x\cdot64+94x\cdot56=22\left(g\right)\)
\(\Rightarrow x=\dfrac{11}{3240}\)
\(n_{H_2}=n_{Fe}=\dfrac{11}{3240}\cdot94=\dfrac{517}{1620}\left(mol\right)\)
\(V_{H_2}=7.15\left(l\right)\)
a)
$FeO + 2HCl \to FeCl_2 + H_2O$
$Fe + 2HCl \to FeCl_2 + H_2$
$2FeO + 4H_2SO_4 \to Fe_2(SO_4)_3 + SO_2 + 4H_2O$
$2Fe + 6H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 + 6H_2O$
b)
n Fe = n H2 = 4,48/22,4 = 0,2(mol)
n SO2 = 7,84/22,4 = 0,35(mol)
Bảo toàn e :
n FeO + 3n Fe = 2n SO2
=> n FeO = 0,35.2 - 0,2.3 = 0,1(mol)
=> m = 0,1.72 + 0,2.56 = 18,4 gam
\(n_{KMnO_4}=0,1\left(mol\right)\)
\(PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(\Rightarrow n_{O_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\)
Đặt \(\left\{{}\begin{matrix}n_{Cu}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)
Bảo toàn e:\(\Rightarrow2a+3b=0,5\)
Mặt khác: \(64a+56b=13,6-0,05.32=12\)
\(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\%m_{Fe}=\dfrac{0,1.56}{12}.100\%=46,67\left(\%\right)\)
Tại sao lại có 2a+3b=0,5 ạ ?
Qúa trình nhường e của Fe diễn ra ntn ạ ?
a, \(Fe+H_2SO_{4\text{loãng}}\rightarrow FeSO_4+H_2\)
\(n_{Fe}=n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(Fe+H_2SO_{4\text{đặc}}\rightarrow Fe_2\left(SO_4\right)_3+SO_2+H_2O\)
\(Cu+H_2SO_{4\text{đặc}}\rightarrow CuSO_4+SO_2+H_2O\)
Bảo toàn e:
\(2n_{Cu}+3n_{Fe}=2n_{SO_2}\)
\(\Leftrightarrow n_{Cu}=\dfrac{2n_{SO_2}-3n_{Fe}}{2}=0,25\left(mol\right)\)
\(\Rightarrow x=m_{Cu}+m_{Fe}=0,25.64+0,5.56=44\left(g\right)\)
a) Đặt \(\left\{{}\begin{matrix}n_{Cu}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)=b=n_{Fe}\\n_{SO_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\end{matrix}\right.\)
Bảo toàn electron: \(2a+3b=2\) \(\Rightarrow2a+3\cdot0,5=2\) \(\Rightarrow a=n_{Cu}=0,25\left(mol\right)\)
\(\Rightarrow x=m_{Cu}+m_{Fe}=0,25\cdot64+0,5\cdot56=44\left(g\right)\)
b) Ta có: \(n_{H_2SO_4\left(p/ư\right)}=\dfrac{1}{2}n_{e\left(traođổi\right)}+n_{SO_2}=\dfrac{1}{2}\cdot2+1=2\left(mol\right)\)
\(\Rightarrow\Sigma n_{H_2SO_4\left(đặc\right)}=2\cdot110\%=2,2\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{2,2\cdot98}{98\%}=220\left(g\right)\) \(\Rightarrow V_{H_2SO_4}=\dfrac{220}{1,84}\approx119,57\left(ml\right)\)
c) Ta có: \(\left\{{}\begin{matrix}n_{SO_2}=1\left(mol\right)\\n_{Ba\left(OH\right)_2}=0,4\cdot1,5=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo 2 muối
PTHH: \(2SO_2+Ba\left(OH\right)_2\rightarrow Ba\left(HSO_3\right)_2\)
2x x x (mol)
\(SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3\downarrow+H_2O\)
y y (mol)
Ta lập được hệ phương trình: \(\left\{{}\begin{matrix}x+y=0,6\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=n_{Ba\left(HSO_3\right)_2}=0,4\left(mol\right)\\y=0,2\end{matrix}\right.\)
\(\Rightarrow C_{M_{Ba\left(HSO_3\right)_2}}=\dfrac{0,4}{0,4}=1\left(M\right)\)
Fe + H2SO4--> FeSO4 + H2
2Fe+ 6H2SO4 đ,n--> Fe2(SO4)3 + 3SO2 + 6H2O(1)
Cu + 2H2SO4 đ, n --> CuSO4 + 2H2O + SO2(2)
Ta có nH2=2,24/22,4=0,1 mol
=> nFe=0,1 mol
=> mCu=12-5,6=6,4 g
=> nCu=6,4/64= 0, 1mol
Theo PTHH ta có nSO2 (1)=3.nFe/2=0,15 mol
nSO2(2)=nCu=0,1 mol
=> VSO2=(0,15+0,1).22,4=5,6 lít
Ta có nFe2(SO4)3=nFe/2=0,05 mol
nCuSO4=nCu=0,1 mol
=>m muối=0,05.400+ 160.0,1=36 g