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\(n_{Zn}=\dfrac{13}{65}=0,2mol\)
\(n_{H_2SO_4}=0,04.1=0,04mol\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,2 > 0,04 ( mol )
0,04 0,04 0,04 0,04 ( mol )
\(m_{ZnSO_4}=0,04.161=6,44g\)
Câu b ko hiểu lắm bạn ơi!
PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
a, Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\). ta được H2SO4 dư.
Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)
b, Ta có: \(m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)
c, \(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,1}{0,2}=0,5M\)
\(C_{M_{ZnSO_4}}=\dfrac{0,1}{0,2}=0,5M\)
Bạn tham khảo nhé!
\(n_{Fe}=\dfrac{11,2}{56}=0,2mol\)
\(n_{H_2SO_4}=0,15.2=0,3mol\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,2 < 0,3 ( mol )
0,2 0,2 0,2 0,2 ( mol )
\(V_{H_2}=0,2.22,4=4,48l\)
\(m_{FeSO_4}=0,2.152=30,4g\)
\(\left\{{}\begin{matrix}C_{M_{FeSO_4}}=\dfrac{0,2}{0,15}=1,33M\\C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,3-0,2}{0,15}=0,67M\end{matrix}\right.\)
\(n_{Fe}=\dfrac{2.8}{56}=0.05\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
a) Chất tan : FeSO4
Chất khí : H2
\(m_{FeSO_4}=0.05\cdot152=7.6\left(g\right)\)
\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0.2..........0.3...............0.1...........0.3\)
\(m_{H_2SO_4}=0.3\cdot98=29.4\left(g\right)\)
\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)
\(m_{dd_{H_2SO_4}}=\dfrac{29.4\cdot100}{20}=147\left(g\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0.1\cdot342=34.2\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng }}=5.4+147-0.3\cdot2=151.8\left(g\right)\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34.2}{151.8}\cdot100\%=22.53\%\)
nAl =5.4275.427=0.2 (mol) đổi 200ml = 0,2l
nH2SO4 = Cm.V =1,35.0,2=0,27(MOL)
2Al + 3H2SO4→→Al2(SO4)3 + 3H2
pt; 2 ; 3 : 1 : 3
đb; 0.18 : 0.27 : 0.09 : 0.27 (mol)
so sánh nAl =0.220.22>nH2SO4 =0.2730.273
a, nAl dư = 0.2-0.18=0.02(mol)
m Al dư = 0,02.27=0.54(g)
b, VHH22=0,27.22,4 = 6,048(l)
c, dd tạo thành sau pư là Al2(SO4)3
Cm Al2(SO4)3 = nVnV=0.090.20.090.2=0.45
nHCl=0,6 mol
FeO+2HCl-->FeCl2+ H2O
x mol x mol
Fe2O3+6HCl-->2FeCl3+3H2O
x mol 2x mol
72x+160x=11,6 =>x=0,05 mol
A/ CFeCl2=0,05/0,3=1/6 M
CFeCl3=0,1/0,3=1/3 M
CHCl du=(0,6-0,4)/0,3=2/3 M
B/
NaOH+ HCl-->NaCl+H2O
0,2 0,2
2NaOH+FeCl2-->2NaCl+Fe(OH)2
0,1 0,05
3NaOH+FeCl3-->3NaCl+Fe(OH)3
0,3 0,1
nNaOH=0,6
CNaOH=0,6/1,5=0,4M
`PTHH: 2Al + 3H_2 SO_4 -> Al_2 (SO_4)_3 + 3H_2↑`
`a) n_[Al] = [ 2,7 ] / 27 = 0,1 (mol)`
`n_[H_2 SO_4] = [ [ 19,6 ] / 100 . 100 ] / 98 = 0,2 (mol)`
Ta có: `[ 0,1 ] / 2 < [0,2] / 3`
`=> H_2 SO_4` dư
Theo `PTHH` có: `n_[H_2 SO_\text{4(p/ứ)}] = 3 / 2 n_[Al] = 3 / 2 . 0,1 = 0,15 (mol)`
`=>m_[H_2 SO_\text{4(dư)}] = ( 0,2 - 0,15 ) . 98 = 4,9 (g)`
_____________________________________________________
`c)`Theo `PTHH` có: `n_[H_2] = 3 / 2 n_[Al] = 3 / 2 . 0,1 = 0,15 (mol)`
`-> m_\text{dd sau p/ứ} = 2,7 + 100 - 0,15 . 2 = 102,5 (g)`
`=> C%_[H_2 SO_\text{4(dư)}] = [ 4,9 ] / [102,5 ] . 100 ~~ 4,78 %`
Theo `PTHH` có: `n_[Al_2 (SO_4)_3] = 1 / 2 n_[Al] = 1 / 2 . 0,1 = 0,05 (mol)`
`=> C%_[Al_2 (SO_4)_3] = [ 0,05 . 342 ] / [ 102,5 ] . 100 ~~ 16,68%`
\(nAl=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(mH_2SO_4=\dfrac{100.19,6}{100}=19,6\left(g\right)\)
\(nH_2SO_4=\dfrac{19,6}{98}=0,2\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
2 3 1 3 (mol)
0,1 0,15 0,05 0,15 (mol)
LTL : \(\dfrac{0,1}{2}< \dfrac{0,2}{3}\)
=> Al đủ , H2SO4 dư
m H2SO4 ( dư ) = ( 0,2 - 0,15 ) . 98 = 4,9 (g)
\(mAl_2\left(SO_4\right)_3=0,05.342=17,1\left(g\right)\)
\(mH_2=0,15.2=0,3\left(g\right)\)
m dd = mAl + mddH2SO4 + mAl2(SO4)3 - mH2
m dd = 2,7 + 100 + 17,1 - 0,3 = 119,5 (g)
\(C\%_{ddH_2SO_4}=\dfrac{4,9.100}{119,5}=4,1\%\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{17,1.100}{119,5}=14,3\%\)
\(C\%_{H_2}=\dfrac{0,3.100}{119,5}=0,25\%\)
nZn=6,5/65=0,1(mol)
nH2SO4=1.0,2=0,2(mol)
Zn+H2SO4--->ZnSO4+H2
1____1
0,1__0,2
Ta có: 0,1/1<0,2/1
=>H2SO4 dư
mH2SO4 dư=0,1.98=9,8(g)
=>CM=mct/mdd=
nFe=1,12/56=0,02(mol)
Fe+H2SO4--->FeSO4+H2
0,02__________0,02
mFeSO4=0,02.152=3,04(g)
C%=3,04/(1,12+200).100%=1,5%