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a, \(FeO+H_2\underrightarrow{t^o}Fe+H_2O\)
b, \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)
\(n_{Fe}=n_{FeO}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
c, \(n_{H_2}=n_{FeO}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(m_{CuO}=50.20\%=10\left(g\right)\)
\(n_{CuO}=\dfrac{10}{80}=0,125\left(mol\right)\)
\(m_{Fe_2O_3}=50-10=40\left(g\right)\)
\(n_{Fe_2O_3}=\dfrac{40}{160}=0,25\left(mol\right)\)
PTHH :
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
0,125 0,125 0,125
\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
0,25 0,75 0,5
\(a,V_{H_2}=\left(0,75+0,125\right).22,4=19,6\left(l\right)\)
\(b,m_{Cu}=0,125.64=8\left(g\right)\)
\(m_{Fe}=0,5.56=28\left(g\right)\)
PT: Fe2O3+3H2to→2Fe+3H2O
CuO+H2to→Cu+H2O
a, Ta có: mFe2O3=20.60%=12(g)
⇒nFe2O3=\(\dfrac{12}{160}\)=0,075(mol
mCuO=20−12=8(g
⇒nCuO=\(\dfrac{8}{80}\)=0,1(mol)
Theo pT:
nFe=2nFe2O3=0,15(mol)
nCu=nCuO=0,1(mol)
⇒mFe=0,15.56=8,4(g)
mCu=0,1.64=6,4(g)
b, Theo PT: nH2=3nFe2O3+nCuO=0,325(mol)
⇒VH2=0,325.22,4=7,28(l)
c. Zn+2HCl->ZnCl2+H2
0,65----------0,325
=>m HCl=0,65.36,5=23,725g
PTHH: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
a+b) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=0,2\left(mol\right)\\n_{H_2}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=0,2\cdot160=32\left(g\right)\\V_{H_2}=0,6\cdot22,4=13,44\left(l\right)\end{matrix}\right.\)
c) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
Theo PTHH: \(n_{Zn}=n_{H_2}=0,6\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,6\cdot65=39\left(g\right)\)
a,
nFe = 22,4/56 = 0,4 (mol)
PTHH
Fe2O3 + 3H2 ---to----) 2Fe + 3H2O (1)
theo phương trình (1) ,ta có:
nFe2O3 = 0,4 x 2 / 1 = 0,8 (mol)
mFe2O3 = 160 x 0,8 = 128 (g)
b,
theo pt (1)
nH2 = (0,4 x 3)/2 = 0,6 (mol)
=) VH2 = 0,6 x 22,4 = 13,44 (L)
c,
PTHH
Zn + H2SO4 -------------) ZnSO4 + H2 (2)
Số mol H2 cần dùng là 0,6 (mol)
Theo PT (2) :
nZn = nH2 ==) nZn = 0,6 x 65 = 39 (g)
a ) \(n_{Fe_2O_4}=\frac{23,2}{232}=0,1\) mol
\(Fe_3O_4+4H_2\underrightarrow{t^0}3Fe+4H_2O\)
0,1 -> 0,4 -> 0,3
\(\Rightarrow n_{H_2}=4n_{Fe_3O_4}=0,4\) mol \(\Rightarrow V_{H_2}=0,4.22,4=8,96\) lít
b ) \(n_{Fe}=3n_{Fe_3O_4}=0,3\) mol \(\Rightarrow m_{Fe}=56.0,3=16,8\) gam.
a ) \(n_{Fe_2O_3}=\frac{32}{160}=0,2\) mol
\(Fe_2O_3+3H_2\underrightarrow{t^0}2Fe+3H_2O\)
0,2 ->0,6 ->0,4
\(\Rightarrow m_{Fe}=56.0,4=22,4\) gam
b ) \(n_{H_2}=3n_{Fe}=0,6\) mol \(\Rightarrow V_{H_2}=0,6.22,4=13,44\) lít .
\(n_{ZnO}=\dfrac{m}{M}=\dfrac{16,2}{65+16}=0,2\left(mol\right)\)
a) \(PTHH:Zn+H_2O\rightarrow ZnO+H_2\)
1 1 1 1
0,2 0,2 0,2 0,2
b) \(V_{H_2}=n.24,79=0,2.24,79=4,958\left(l\right)\)
c) \(m_{Zn}=n.M=0,2.65=13\left(g\right).\)
\(n_{Fe_2O_3}=\dfrac{14.4}{160}=0.09\left(mol\right)\)
\(Fe_2O_3+3H_2\underrightarrow{^{t^0}}2Fe+3H_2O\)
\(0.09.........0.27...0.18\)
\(V_{H_2}=0.27\cdot22.4=6.048\left(l\right)\)
\(m_{Fe}=0.18\cdot56=10.08\left(g\right)\)
Bn có thể lm rõ hơn đc không