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a.
Ta có:
nFe=2,8\56=0,05mol
Phương trình hóa học
Fe+2HCl→FeCl2+H2
0,05→0,1mol
→VHCl=n\CM=0,12=0,05l
b.
nH2=nFe=0,05mol
→VH2=0,05.22,4=1,12l
c.
nFeCl2=nFe=0,05mol
→CMFeCl2=0,05\0,05=1M
a) \(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,05 0,1 0,05 0,05
b) \(V_{ddHCl}=\dfrac{0,1}{2}=0,05\left(l\right)=50\left(ml\right)\)
c) \(V_{H_2}=0,05.24,79=1,2395\left(l\right)\)
d) \(C_{M_{ddFeCl_2}}=\dfrac{0,05}{0,05}=1M\)
a) 250ml=0,25L
\(n_{FeSO_4\left(1,2M\right)}=v\cdot C_M=0,25\cdot1,2=0,3\left(mol\right)\)
\(n_{FeSO_4\left(1,5M\right)}=v\cdot C_M=0,25\cdot1,5=0,375\left(mol\right)\)
b) \(n_{FeSO_4}=n_{FeSO_4\left(1,5M\right)}-n_{FeSO_4\left(1,2M\right)}=0,375-0,3=0,075\left(mol\right)\)
\(m_{FeSO_4}=n\cdot M=0,075\cdot152=11,4\left(g\right)\)
\(n_{Zn}=\dfrac{13}{65}=0,2mol\)
\(n_{H_2SO_4}=0,04.1=0,04mol\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
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!
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\left(1\right)\)
x x
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\left(2\right)\)
y 1,5y
Đổi: 400ml = 0,4l
\(n_{H_2SO_4}=C_M.V=1.0,4=0,4\left(mol\right)\)
Gọi x,y lần lượt là số mol của Fe, Al
Từ(1),(2) =>\(\left\{{}\begin{matrix}x+1,5y=0,4\\56x+27y=11\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(m_{Al}=0,2.27=5,4\left(g\right)\)
\(m_{Fe}=0,1.56=5,6\left(g\right)\)
\(C_{M_{FeSO_4}}=\dfrac{0,1}{0,4}=0,25M\)
\(C_{M_{Al_2\left(so_4\right)_3}}=\dfrac{0,1}{0,4}=0,25M\)
a, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=0,2.1,35=0,27\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{2}>\dfrac{0,27}{3}\), ta được Al dư.
Theo PT: \(n_{H_2}=n_{H_2SO_4}=0,27\left(mol\right)\Rightarrow V_{H_2}=0,27.22,4=6,048\left(l\right)\)
b, \(n_{Al\left(pư\right)}=\dfrac{2}{3}n_{H_2SO_4}=0,18\left(mol\right)\)
\(\Rightarrow m_{Al\left(pư\right)}=0,18.27=4,86\left(g\right)\)
c, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2SO_4}=0,09\left(mol\right)\)
\(\Rightarrow C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,09}{0,2}=0,45\left(M\right)\)
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
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!
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{H_2SO_4}=0,5.0,1=0,05\left(mol\right)\\ pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(LTL:\dfrac{0,2}{2}>\dfrac{0,05}{3}\)
=> Al dư
\(n_{H_2}=n_{H_2SO_4}=0,05\left(mol\right)\)
\(V_{H_2}=0,05.22,4=1,12l\\
C_M=\dfrac{\dfrac{1}{60}}{0,1}=0,16M\)
Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=0,1\left(mol\right)\\n_{NaOH}=0,015\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,015}{0,1}=0,15\)
=> Tạo muối NaHCO3
\(CO_2\left(0,015\right)+NaOH\left(0,015\right)\rightarrow NaHCO_3\left(0,015\right)\)
\(\Rightarrow C_{M_{NaHCO_3}}=\dfrac{0,015}{0,015}=1\left(M\right)\)